Trigonometric Ratios And Identities

(i) Sin C

We know that,

So, here θ = C

Side opposite to ∠C = AB = 3

Hypotenuse = AC = 5

So,

(ii) Sin A

So, here θ = A

The side opposite to ∠A = BC = 4

Hypotenuse = AC = 5

So,

(iii) Cos C

We know that,

So, here θ = C

Side adjacent to ∠C = BC = 4

Hypotenuse = AC = 5

So,

(iv) Cos A

Here, θ = A

Side adjacent to ∠A = AB = 3

Hypotenuse = AC = 5

So,

(v) tan C

We know that,

So, here θ = C

Side opposite to ∠C = AB = 3

Side adjacent to ∠C = BC = 4

So,

(vi) tan A

here θ = A

Side opposite to ∠A = BC = 4

Side adjacent to ∠A = AB = 3

So,

(i) tan θ

We know that,

Side opposite to θ = AB = 4

Side adjacent to θ = BC = 3

So,

(ii) cos θ

We know that,

Side adjacent to θ = BC = 3

Hypotenuse = AC = 5

So,

(i) sin θ

We know that,

Side opposite to θ = BC = ?

Hypotenuse = AC = 13

Firstly we have to find the value of BC.

So, we can find the value of BC with the help of Pythagoras theorem.

According to Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

⇒ (AB)² + (BC)² = (AC)²

⇒ (12)² + (BC)² = (13)²

⇒ 144 + (BC)² = 169

⇒ (BC)² = 169–144

⇒ (BC)² = 25

⇒ BC =√25

⇒ BC =±5

But side BC can’t be negative. So, BC = 5

Now, BC = 5 and AC = 13

So,

(ii) tan θ

We know that,

Side opposite to θ = BC = 5

Side adjacent to θ = AB = 12

So,

(iii) tan A – cot C

We know that,

and

tan A

Here, θ = A

Side opposite to ∠A = BC = 5

Side adjacent to ∠A = AB = 12

So,

Cot C

Here, θ = C

Side adjacent to ∠C = BC = 5

Side opposite to ∠C = AB = 12

So,

So,

(i)

(a) sin A

We know that,

So, here θ = A

Side opposite to ∠A = BC = 7

Hypotenuse = AC = ?

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem.

According to Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

⇒ (AB)² + (BC)² = (AC)²

⇒ (24)² + (7)² = (AC)²

⇒ 576 + 49 = (AC)²

⇒ (AC)² = 625

⇒ AC =√625

⇒ AC =±25

But side AC can’t be negative. So, AC = 25cm

Now, BC = 7 and AC = 25

So,

Cos A

We know that,

So, here θ = A

Side adjacent to ∠A = AB = 24

Hypotenuse = AC = 25

So,

(b) sin C

We know that,

So, here θ = C

The side opposite to ∠C = AB = 24

Hypotenuse = AC = 25

So,

Cos C

We know that,

So, here θ = C

Side adjacent to ∠C = BC = 7

Hypotenuse = AC = 25

So,

(a) Cos²θ +sin² θ

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem.

According to Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

⇒ (AC)² + (BC)² = (AB)²

⇒ (AC)² + (21)² = (29)²

⇒ (AC)² = (29)² – (21)²

Using the identity a² –b² = (a+b) (a – b)

⇒ (AC)² = (29–21)(29+21)

⇒ (AC)² = (8)(50)

⇒ (AC)² = 400

⇒ AC =√400

⇒ AC =±20

But side AC can’t be negative. So, AC = 20units

Now, we will find the sin θ and cos θ

In ∆ACB, Side opposite to angle θ = AC = 20

and Hypotenuse = AB = 29

So,

Now, We know that

In ∆ACB, Side adjacent to angle θ = BC = 21

and Hypotenuse = AB = 29

So,

So

=1

Cos²θ +sin² θ = 1

(b) Cos²θ – sin² θ

Putting values, we get

Given that ∠A is a right angle.

(a) AB = 12, AC = 5, BC = 13

To Find : sin B, cos C and tan B

We know that,

Here, θ = B

Side opposite to angle B = AC = 5

Hypotenuse = BC =13

So,

Now, Cos C

We know that,

Here, θ = C

Side adjacent to angle C = AC = 5

Hypotenuse = BC =13

So,

Now, tan B

We know that,

Here, θ = B

The side opposite to angle B = AC = 5

The side adjacent to angle B = AB = 12

So,

(b) AB = 20, AC = 21, BC = 29

To Find: sin B, cos C and tan B

We know that,

Here, θ = B

The side opposite to angle B = AC =21

Hypotenuse = BC =29

So,

Now, Cos C

We know that,

Here, θ = C

Side adjacent to angle C = AC = 21

Hypotenuse = BC = 29

So,

Now, tan B

We know that,

Here, θ = B

The side opposite to angle B = AC = 21

The side adjacent to angle B = AB = 20

So,

(c) BC =√2, AB = AC = 1

To Find: sin B, cos C and tan B

We know that,

Here, θ = B

The side opposite to angle B = AC =1

Hypotenuse = BC =√2

So

Now, Cos C

We know that,

Here, θ = C

Side adjacent to angle C = AC = 1

Hypotenuse = BC = √2

So,

Now, tan B

We know that,

Here, θ = B

The side opposite to angle B = AC = 1

The side adjacent to angle B = AB = 1

So,

Firstly, we give the name to the midpoint of BC i.e. M

BC = BM + MC = 2BM or 2MC

⇒ BM = 5 and MC = 5

Now, we have to find the value of AM, and we can find out with the help of Pythagoras theorem.

So, In ∆AMB

⇒ (AM)² + (BM)² = (AB)²

⇒ (AM)² + (5)² = (13)²

⇒ (AM)² = (13)² – (5)²

Using the identity a² –b² = (a+b) (a – b)

⇒ (AM)² = (13–5)(13+5)

⇒ (AM)² = (8)(18)

⇒ (AM)² = 144

⇒ AM =√144

⇒ AM =±12

But side AM can’t be negative. So, AM = 12

a. sin θ

We know that,

In ∆AMB

Side opposite to θ = AM = 12

Hypotenuse = AB=13

So,

So,

b. cos θ

We know that,

In ∆AMB

The side adjacent to θ = BM = 5

Hypotenuse = AB = 13

So,

So,

c. tan θ

We know that,

In ∆AMB

Side opposite to θ = AM = 12

The side adjacent to θ = BM = 5

So,

So,

Firstly, we have to find the value of XM and we can find out with the help of Pythagoras theorem

So, In ∆XMZ

⇒ (XM)² + (MZ)² = (XZ)²

⇒ (XM)² + (16)² = (20)²

⇒ (XM)² = (20)² – (16)²

Using the identity a² –b² = (a+b) (a – b)

⇒ (XM)² = (20–16)(20+16)

⇒ (XM)² = (4)(36)

⇒ (XM)² = 144

⇒ XM =√144

⇒ XM =±12

But side XM can’t be negative. So, XM = 12

Now, In ∆XMY we have the value of XM and MY but we don’t have the value of XY.

So, again we apply the Pythagoras theorem in ∆XMY

⇒ (XM)² + (MY)² = (XY)²

⇒ (12)² + (5)² = (XY)²

⇒ 144 + 25 = (XY)²

⇒ (XY)² = 169

⇒ XY =√169

⇒ XY =±13

But side XY can’t be negative. So, XY = 13

a. sin θ

We know that,

In ∆XMY

Side opposite to θ = MY = 5

Hypotenuse = XY = 13

So,

b. cos θ

We know that,

In ∆XMY

Side adjacent to θ = XM = 12

Hypotenuse = XY = 13

So

c. tan θ

We know that,

In ∆XMY

The side opposite to θ = MY = 5

Side adjacent to θ = XM = 12

So,

Given : PQ = 3, QR = 4

⇒ (PQ)² + (QR)² = (PR)²

⇒ (3)² + (4)² = (PR)²

⇒ 9 + 16 = (PR)²

⇒ (PR)² = 25

⇒ PR =√25

⇒ PR =±5

But side PR can’t be negative. So, PR = 5

(i) sin α

We know that,

Here, θ = α

The side opposite to angle α = QR =4

Hypotenuse = PR =5

So,

(ii) cos α

We know that,

Here, θ = α

The side adjacent to angle α = PQ =3

Hypotenuse = PR =5

So,

(iii) tan α

We know that,

Here, θ = α

Side opposite to angle α = QR =4

Side adjacent to angle α = PQ =3

So,

(iv) sin β

We know that,

Here, θ = β

The side opposite to angle β = PQ =3

Hypotenuse = PR =5

So,

(v) cos β

We know that,

Here, θ = β

Side adjacent to angle β = QR =4

Hypotenuse = PR =5

So,

(vi) tan β

We know that,

Here, θ = β

Side opposite to angle β = PQ =3

Side adjacent to angle β = QR =4

So,

Given:

We know that,

Or

Let,

Perpendicular =AB =4k

and Hypotenuse =AC =5k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

In right angled ∆ ABC, we have

⇒ (AB)² + (BC)² = (AC)²

⇒ (4k)² + (BC)² = (5k)²

⇒ 16k² + (BC)² = 25k²

⇒ (BC)² = 25 k² –16 k²

⇒ (BC)² = 9 k²

⇒ BC =√9 k²

⇒ BC =±3k

But side BC can’t be negative. So, BC = 3k

Now, we have to find the value of cos θ and tan θ

We know that,

The side adjacent to angle θ or base = BC =3k

Hypotenuse = AC =5k

So,

Now,

We know that,

Perpendicular = AB =4k

Base = BC =3k

So,

Given: Sin A

We know that,

Or

Let,

Side opposite to angle θ = BC =3k

and Hypotenuse = AC =4k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (AB)² + (BC)² = (AC)²

⇒ (AB)² + (3k)² = (4k)²

⇒ (AB)² + 9k² = 16k²

⇒ (AB)² = 16 k² – 9 k²

⇒ (AB)² = 7 k²

⇒ AB =k√7

So, AB = k√7

Now, we have to find the value of cos A and tan A

We know that,

Here, θ = A

The side adjacent to angle A = AB =k√7

Hypotenuse = AC =4k

So,

Now,

We know that,

The side opposite to angle A = BC =3k

The side adjacent to angle A = AB =k√7

So,

Given:

We know that,

Or

Let,

Perpendicular =AB =3k

and Hypotenuse =AC =5k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (AB)² + (BC)² = (AC)²

⇒ (3k)² + (BC)² = (5k)²

⇒ 9k² + (BC)² = 25k²

⇒ (BC)² = 25 k² – 9 k²

⇒ (BC)² = 16 k²

⇒ BC =√16 k²

⇒ BC =±4k

But side BC can’t be negative. So, BC = 4k

Now, we have to find the value of cos θ and tan θ

We know that,

The side adjacent to angle θ = BC =4k

Hypotenuse = AC =5k

So,

Now, tan θ

We know that,

Perpendicular = AB =3k

Base = BC =4k

So,

We know that,

Or

Let,

Base =BC = 4k

Hypotenuse =AC = 5k

Where, k ia any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (AB)² + (BC)² = (AC)²

⇒ (AB)² + (4k)² = (5k)²

⇒ (AB)² + 16k² = 25k²

⇒ (AB)² = 25 k² –16 k²

⇒ (AB)² = 9 k²

⇒ AB =√9 k²

⇒ AB =±3k

But side AB can’t be negative. So, AB = 3k

Now, we have to find tan θ

We know that,

Side opposite to angle θ = BC =4k

Side adjacent to angle θ = AB =3k

So

We know that,

Or

Let,

The side opposite to angle θ =AB = 3k

The side adjacent to angle θ =BC = 4k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (3k)² + (4k)² = (AC)²

⇒ (AC)² = 9 k²+16 k²

⇒ (AC)² = 25 k²

⇒ AC =√25 k²

⇒ AC =±5k

But side AC can’t be negative. So, AC = 5k

Now, we will find the sin θ and cos θ

Side opposite to angle θ = AB = 3k

and Hypotenuse = AC = 5k

So,

Now, We know that

Side adjacent to angle θ = BC = 4k

and Hypotenuse = AC = 5k

So,

We know that,

Or

Here, θ = A

Let,

The side opposite to angle A =BC = 4k

The side adjacent to angle A =AB = 3k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (3k)² + (4k)² = (AC)²

⇒ (AC)² = 9 k² +16 k²

⇒ (AC)² = 25 k²

⇒ AC =√25 k²

⇒ AC =±5k

But side AC can’t be negative. So, AC = 5k

Now, we will find the sin A and cos A

Side opposite to angle A = BC = 4k

and Hypotenuse = AC = 5k

So,

Now, We know that

Side adjacent to angle A = AB = 3k

and Hypotenuse = AC = 5k

So,

Now, we find other trigonometric ratios

We know that,

Or

Let,

Side adjacent to angle θ =AB = 12k

The side opposite to angle θ =BC = 5k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (12k)² + (5k)² = (AC)²

⇒ (AC)² = 144 k² +25 k²

⇒ (AC)² = 169 k²

⇒ AC =√169 k²

⇒ AC =±13k

But side AC can’t be negative. So, AC = 13k

Now, we will find the sin θ

Side opposite to angle θ = BC = 5k

and Hypotenuse = AC = 13k

So,

We know that,

Or

Let,

The side opposite to angle θ =BC = 5k

The side adjacent to angle θ =AB = 12k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (12k)² + (5k)² = (AC)²

⇒ (AC)² = 144 k² +25 k²

⇒ (AC)² = 169 k²

⇒ AC =√169 k²

⇒ AC =±13k

But side AC can’t be negative. So, AC = 13k

Now, We know that

Side adjacent to angle θ = AB = 12k

and Hypotenuse = AC = 13k

So,

Given: Sin θ

We know that,

Or

Let,

Side opposite to angle θ = 12k

and Hypotenuse = 13k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (AB)² + (BC)² = (AC)²

⇒ (12k)² + (BCk)² = (13)²

⇒ 144 k² + (BC)² = 169 k²

⇒ (BC)² = 169 k² –144 k²

⇒ (BC)² = 25 k²

⇒ BC =√25 k²

⇒ BC =±5k

But side BC can’t be negative. So, BC = 5k

Now, we have to find the value of cos θ and tan θ

We know that,

Side adjacent to angle θ = BC =5k

Hypotenuse = AC =13k

So,

Now, tan θ

We know that,

side opposite to angle θ = AB =12k

Side adjacent to angle θ = BC =5k

So,

We know that,

Or

Given: tan θ =0.75

Let,

The side opposite to angle θ =BC = 3k

The side adjacent to angle θ =AB = 4k

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (4k)² + (3k)² = (AC)²

⇒ (AC)² = 16 k² +9 k²

⇒ (AC)² = 25 k²

⇒ AC =√25 k²

⇒ AC =±5k

But side AC can’t be negative. So, AC = 5k

Now, we will find the sin θ

Side opposite to angle θ = BC = 3k

and Hypotenuse = AC = 5k

So,

We know that,

Or

Given: tan B = √3

Let,

Side opposite to angle B =AC = √3k

The side adjacent to angle B =AB = 1k

where k is any positive integer

Firstly we have to find the value of BC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (AC)² = (BC)²

⇒ (1k)² + (√3k)² = (BC)²

⇒ (BC)² = 1 k² +3 k²

⇒ (BC)² = 4 k²

⇒ BC =√2 k²

⇒ BC =±2k

But side BC can’t be negative. So, BC = 2k

Now, we will find the sin B and cos B

Side opposite to angle B = AC = k√3

and Hypotenuse = BC = 2k

So,

Now, we know that,

The side adjacent to angle B = AB =1k

Hypotenuse = BC =2k

So,

We know that,

Or

Here,

So, Side opposite to angle θ =AC = m

The side adjacent to angle θ =AB = n

Firstly we have to find the value of BC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (AC)² = (BC)²

⇒ (n)² + (m)² = (BC)²

⇒ (BC)² = m² + n²

⇒ BC =√ m² + n²

So, BC =√(m² + n²)

Now, we will find the sin B and cos B

Side opposite to angle θ = AC = m

and Hypotenuse = BC =√(m² + n²)

So,

Now, we know that,

Side adjacent to angle θ = AB =n

Hypotenuse = BC =√(m² + n²)

So,

Given : sin θ =√3cos θ

⇒ tan θ =√3

We know that,

Or

and tan θ = √3

Let,

The side opposite to angle θ =AC = k√3

The side adjacent to angle θ =AB = 1k

where k is any positive integer

Firstly we have to find the value of BC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (AC)² = (BC)²

⇒ (1k)² + (k√3)² = (BC)²

⇒ (BC)² = 1 k² +3 k²

⇒ (BC)² = 4 k²

⇒ BC =√2 k²

⇒ BC =±2k

But side BC can’t be negative. So, BC = 2k

Now, we will find the sin θ and cos θ

Side opposite to angle θ = AC = k√3

and Hypotenuse = BC = 2k

So,

Now, we know that,

The side adjacent to angle θ = AB =1k

Hypotenuse = BC =2k

So,

We know that,

Or

Let,

Side adjacent to angle θ =AB = 21k

The side opposite to angle θ =BC = 20k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (21k)² + (20k)² = (AC)²

⇒ (AC)² = 441 k² +400 k²

⇒ (AC)² = 841 k²

⇒ AC =√841 k²

⇒ AC =±29k

But side AC can’t be negative. So, AC = 29k

Now, we will find the sin θ

Side opposite to angle θ = BC = 20k

and Hypotenuse = AC = 29k

So,

Now, we know that,

Side adjacent to angle θ = AB =21k

Hypotenuse = AC =29k

So,

Given: 15 cot A = 8

And we know that,

Or

Let,

Side adjacent to angle A =AB = 8k

The side opposite to angle A =BC = 15k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (8k)² + (15k)² = (AC)²

⇒ (AC)² = 64 k² +225 k²

⇒ (AC)² = 289 k²

⇒ AC =√289 k²

⇒ AC =±17k

But side AC can’t be negative. So, AC = 17k

Now, we will find the sin θ

Side opposite to angle θ = BC = 15k

and Hypotenuse = AC = 17k

So,

Now, we know that,

The side adjacent to angle θ = AB =8

Hypotenuse = AC =17

So,

Given: sinθ = cosθ

⇒ tan θ = 1

Let,

Side opposite to angle θ = AB =1k

The side adjacent to angle θ = BC =1k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (1k)² + (1k)² = (AC)²

⇒ (AC)² = 1k² +1k²

⇒ (AC)² = 2k²

⇒ AC =√2k²

⇒ AC =k√2

So, AC = k√2

Now, we will find the sin θ

Side opposite to angle θ = AB= 1k

and Hypotenuse = AC = k√2

So,

Now, we know that,

The side adjacent to angle θ = BC =1k

Hypotenuse = AC =k√2

So,

We know that,

Or,

Let,

Side opposite to angle θ = AB = x² – y²

and Hypotenuse = AC = x² + y²

In right angled ∆ABC, we have

(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]

⇒ (x² – y² )² + (BC)² = (x² + y² )²

⇒ (BC)² = (x² + y² )² – (x² – y² )²

Using the identity, a² – b² = (a+b)(a – b)

⇒ (BC)² = [(x² + y² + x² – y² )][ x² + y² –( x² – y²)]

⇒ (BC)² = (2x²)(2y²)

⇒ (BC)² = (4x²y²)

⇒ BC =√4x²y²

⇒ BC = ±2xy

⇒ BC = 2xy [taking positive square root since, side cannot be negative]

and

So,

Given:

We know that,

Let,

AB = √(m² – n²) and BC = n

In right angled ∆ABC, we have

(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]

⇒ (√(m² – n²))² + (n)² = (AC )²

⇒ m² – n² + n² = (AC )²

⇒ (AC)² = (m²)

⇒ AC =√ m²

⇒ AC = ±m

⇒ AC = m [taking positive square root since, side cannot be negative]

Now, we have to find the value of cos θ and sin θ

We, know that

and

Given: sec θ = 2

We know that,

Let,

BC = 1k and AC = 2k

where, k is any positive integer.

In right angled ∆ABC, we have

(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]

⇒ (AB)² + (1k)² = (2k )²

⇒ (AB)² + k² = 4k²

⇒ (AB)² = 4k² – k²

⇒ (AB)² = 3k²

⇒ AB = k√3

Now, we have to find the value of other trigonometric ratios.

We, know that

Given:

We know that,

Let,

BC = 12k and AC = 13k

where, k is any positive integer.

In right angled ∆ABC, we have

(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]

⇒ (AB)² + (12k)² = (13k )²

⇒ (AB)² + 144k² = 169k²

⇒ (AB)² = 169k² – 144k²

⇒ (AB)² = 25k²

⇒ AB = √25k²

⇒ AB =±5k [taking positive square root since, side cannot be negative]

Now, we have to find the value of other trigonometric ratios.

We, know that

Given: cosec θ = √10

We know that,

Let,

AB = 1k and AC = k√10

where, k is any positive integer.

In right angled ∆ABC, we have

(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]

⇒ (1k )²+ (BC)² = (k√10)²

⇒ (BC)² = 10k² – k²

⇒ (BC)² = 9k²

⇒ BC = √9k²

⇒ BC =±3k [taking positive square root since, side cannot be negative]

Now, we have to find the value of other trigonometric ratios.

We, know that

We know that,

Or

Given:

Let,

Side opposite to angle A =BC = k√3

Side adjacent to angle A =AB = 2k

where, k is any positive integer

Firstly we have to find the value of BC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (2k)² + (√3k)² = (AC)²

⇒ (AC)² = 4 k² +3 k²

⇒ (AC)² = 7 k²

⇒ AC =√7 k²

⇒ AC =k√7

So, AC = k√7

Now, we will find the sin A and cos A

Side opposite to angle A = BC = k√3

and Hypotenuse = AC = k√7

So,

Now, we know that,

Side adjacent to angle A = AB =2k

Hypotenuse = AC = k√7

So,

Now, we have to find sin A +cos A

Putting values of sin A and cos A, we get

Given: sin θ =√3cos θ

⇒ tan θ = √3

We know that,

Or

Given: tan θ = √3

Let,

Side opposite to angle θ =AC = √3k

Side adjacent to angle θ =AB = 1k

where k is any positive integer

Firstly we have to find the value of BC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (AC)² = (BC)²

⇒ (1k)² + (√3k)² = (BC)²

⇒ (BC)² = 1 k² +3 k²

⇒ (BC)² = 4 k²

⇒ BC =√2 k²

⇒ BC =±2k

But side BC can’t be negative. So, BC = 2k

Now, we will find the sin B and cos B

Side opposite to angle θ = AC = k√3

and Hypotenuse = BC = 2k

So,

Now, we know that,

The side adjacent to angle θ = AB =1k

Hypotenuse = BC =2k

So,

Now, we have to find the value of cos θ – sin θ

Putting the values of sin θ and cos θ, we get

We know that,

Or

Let,

Side opposite to angle θ =AB = 8k

Side adjacent to angle θ =BC = 15k

where, k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (8k)² + (15k)² = (AC)²

⇒ (AC)² = 64k²+225k²

⇒ (AC)² = 289 k²

⇒ AC =√289 k²

⇒ AC =±17k

But side AC can’t be negative. So, AC = 17k

Now, we will find the cos θ

We know that

Side adjacent to angle θ = BC = 15k

and Hypotenuse = AC = 17k

So,

Now, we have to find the value of 1+ cos² θ

Putting the value of cos θ, we get

Given:

We know that,

Or

Let,

Side adjacent to angle θ =AB = 7k

Side opposite to angle θ =BC = 8k

where, k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (7k)² + (8k)² = (AC)²

⇒ (AC)² = 49 k² +69 k²

⇒ (AC)² = 113 k²

⇒ AC =√113 k²

⇒ AC =k√113

(i)

We know that,

(a+b)(a – b) = (a² – b²)

So, using this identity, we get

(ii) cot² θ

Given

Given: 3cot A = 4

We know that,

Or

Let,

Side adjacent to angle A =AB = 4k

The side opposite to angle A =BC = 3k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (4k)² + (3k)² = (AC)²

⇒ (AC)² = 16 k² + 9 k²

⇒ (AC)² = 25 k²

⇒ AC =√25k²

⇒ AC = ±5k [taking positive square root since, side cannot be negative]

and

Now,

…(i)

And RHS = cos² A – sin² A

=

…(ii)

From Eqs. (i) and (ii) LHS =RHS

Hence Proved

tan A = 1

As we know

Now construct a right angle triangle right angled at B such that

∠ BAC = θ

Hence perpendicular = BC = 1 and base = AB = 1

By Pythagoras theorem,

AC² = AB² + BC²

⇒ AC² = (1)² + (1)²

⇒ AC² = 2

⇒ AC =

As,

⇒

Hence,

2 sin A cos A=

⇒ 2 sin A cos A=

⇒ 2 sin A cos A=1

= R.H.S

Hence proved.

4sin² θ =3

But it is given 0^o< θ <90^o

So,

Let, P =k√3 and H =2k

In right angled ∆ABC, we have

B² + P² = H²

⇒ B² + (k√3)² = (2k)²

⇒ B² + 3k² = 4k²

⇒ B² = 4k² – 3k²

⇒ B² = k²

⇒ B = ±k

⇒ B = k [taking positive square root since, side cannot be negative]

So,

Given:

Now,

Given: 13 cosθ = 5

We know that,

Let AB =5k and BC = 13k

In right angled ∆ABC, we have

B² + P² = H²

⇒ (5k)² + P² = (13k)²

⇒ P² + 25k² = 169k²

⇒ P² = 169k² – 25k²

⇒ P² = 144k²

⇒ P =√144k²

⇒ P = ±12k

⇒ P = 12k [taking positive square root since, side cannot be negative]

Now,

Given:

We know that,

Let,

BC = 5k and AC = 13k

where, k is any positive integer.

In right angled ∆ABC, we have

(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]

⇒ (AB)² + (5k)² = (13k )²

⇒ (AB)² + 25k² = 169k²

⇒ (AB)² = 169k² – 25k²

⇒ (AB)² = 144k²

⇒ AB = √144k²

⇒ AB =±12k [taking positive square root since, side cannot be negative]

Now, we have to find the value of other trigonometric ratios.

We, know that

Now, LHS =

=3 =RHS

Hence Proved

Given: 2 tan θ = 1

We know that,

Or

Let,

Side opposite to angle θ =AB = 1k

Side adjacent to angle θ =BC = 2k

where, k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (k)² + (2k)² = (AC)²

⇒ (AC)² = k²+4k²

⇒ (AC)² = 5k²

⇒ AC =√5k²

⇒ AC =±k√5

But side AC can’t be negative. So, AC = k√5

Now, we will find the sin θ and cos θ

We know that

Side adjacent to angle θ = BC = 2k

and Hypotenuse = AC = k√5

So,

And

Side adjacent to angle θ =AB = 1k

And Hypotenuse =AC = k√5

So,

Now,

Given: 5 tan = 4

⇒ tan α

We know that,

Or

Let,

The side opposite to angle α =AB = 4k

The side adjacent to angle α =BC = 5k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (4k)² + (5k)² = (AC)²

⇒ (AC)² = 16k²+25k²

⇒ (AC)² = 41k²

⇒ AC =√41k²

⇒ AC =±k√41

But side AC can’t be negative. So, AC = k√41

Now, we will find the sin α and cos α

We know that

Side adjacent to angle α = BC = 5k

and Hypotenuse = AC = k√41

So,

And

Side adjacent to angle α =AB = 4k

And Hypotenuse =AC = k√5

So,

Now, LHS

= RHS

Hence Proved

We know that,

Or

Let,

Side adjacent to angle θ =AB = 3k

The side opposite to angle θ =BC = 4k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (3k)² + (4k)² = (AC)²

⇒ (AC)² = 9k² +16k²

⇒ (AC)² = 25k²

⇒ AC =√25k²

⇒ AC =±5k

But side AC can’t be negative. So, AC = 5k

Now, we will find the sin θ

Side opposite to angle θ = BC = 4k

and Hypotenuse = AC = 5k

So,

Now, we know that,

The side adjacent to angle θ = AB =3k

Hypotenuse = AC =5k

So,

And

Now, LHS

= RHS

Hence Proved

We know that,

Or

Let,

Side adjacent to angle θ =AB = 1k

Side opposite to angle θ =AC = k√3

where, k is any positive integer

Firstly we have to find the value of BC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (AC)² = (BC)²

⇒ (1k)² + (√3k)² = (BC)²

⇒ (BC)² = 1 k² +3 k²

⇒ (BC)² = 4 k²

⇒ BC =√2 k²

⇒ BC =±2k

But side BC can’t be negative. So, BC = 2k

Now, we will find the sin θ and cos θ

Side opposite to angle θ = AC = k√3

and Hypotenuse = BC = 2k

So

Now, we know that,

Side adjacent to angle θ = AB =1k

Hypotenuse = BC =2k

So,

Now, LHS

= RHS

Hence Proved

We know that,

Or

Let,

Side opposite to angle θ =AB = x

Side adjacent to angle θ =BC = y

where, k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (x)² + (y)² = (AC)²

⇒ (AC)² = x²+y²

⇒ AC =√( x²+y²)

Now, we will find the sin θ and cos θ

We know that

Side adjacent to angle θ = BC = y

and Hypotenuse = AC = √( x²+y²)

So,

And

Side adjacent to angle θ =AB = x

And Hypotenuse =AC = √( x²+y²)

So,

Now, x sin θ +y cos θ

= √( x²+y²)

Given: Sin θ

We know that,

Or

Let,

Perpendicular =AB =3k

and Hypotenuse =AC =5k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

In right angled ∆ ABC, we have

⇒ (AB)² + (BC)² = (AC)²

⇒ (3k)² + (BC)² = (5k)²

⇒ 9k² + (BC)² = 25k²

⇒ (BC)² = 25 k² –9k²

⇒ (BC)² = 16k²

⇒ BC =√16k²

⇒ BC =±4k

But side BC can’t be negative. So, BC = 4k

Now, we have to find the value of cos θ and tan θ

We know that,

The side adjacent to angle θ or base = BC =4k

Hypotenuse = AC =5k

So,

Now,

We know that,

Perpendicular = AB =3k

Base = BC =4k

So,

Now, tan² θ + sin θ cos θ + cot θ

Given: cot θ

We know that,

Or

Let,

Side adjacent to angle θ =AB = 3k

The side opposite to angle θ =BC = 4k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (3k)² + (4k)² = (AC)²

⇒ (AC)² = 9k² +16k²

⇒ (AC)² = 25k²

⇒ AC =√25k²

⇒ AC =±5k

But side AC can’t be negative. So, AC = 5k

Now, we will find the sin θ

Side opposite to angle θ = BC = 4k

and Hypotenuse = AC = 5k

So,

Now, we know that,

Side adjacent to angle θ = AB =3k

Hypotenuse = AC =5k

So,

Now, LHS

= 7 = RHS

Hence Proved

Given: Sin θ

We know that,

Or

Let,

Perpendicular =AB =m

and Hypotenuse =AC =√(m² + n²)

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

In right angled ∆ ABC, we have

⇒ (AB)² + (BC)² = (AC)²

⇒ (m)² + (BC)² = (√(m² + n²))²

⇒ m² + (BC)² = m² + n²

⇒ (BC)² = m² + n² – m²

⇒ (BC)² = n²

⇒ BC =√n²

⇒ BC =±n

But side BC can’t be negative. So, BC = n

Now, we have to find the value of cos θ and tan θ

We know that,

Side adjacent to angle θ or base = BC =n

Hypotenuse = AC =√(m² + n²)

So,

Now, LHS = m sin θ +n cosθ

=√(m² + n²) = RHS

Hence Proved

We know that,

Or

Let,

Base =BC = 12k

Hypotenuse =AC = 13k

Where, k ia any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (AB)² + (BC)² = (AC)²

⇒ (AB)² + (12k)² = (13k)²

⇒ (AB)² + 144k² = 169k²

⇒ (AB)² = 169 k² –144 k²

⇒ (AB)² = 25 k²

⇒ AB =√25 k²

⇒ AB =±5k

But side AB can’t be negative. So, AB = 5k

Now, we have to find sin α and tan α

We know that,

Side opposite to angle α = AB =5k

And Hypotenuse = AC =13k

So,

We know that,

Side opposite to angle α = AB =5k

Side adjacent to angle α = BC =12k

So,

Now, LHS = sin α (1 – tan α)

= RHS

Hence Proved

Given : q cos θ = √(q² – p²)

We know that,

Or

Let,

Base =BC = √(q² – p²)

Hypotenuse =AC = q

Where, k ia any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (AB)² + (BC)² = (AC)²

⇒ (AB)² + (√(q² – p²))² = (q)²

⇒ (AB)² + (q² – p²) = q²

⇒ (AB)² = q² – q² + p²)

⇒ (AB)² = p²

⇒ AB =√p²

⇒ AB =±p

But side AB can’t be negative. So, AB = p

Now, we have to find sin θ

We know that,

The side opposite to angle θ = AB =p

And Hypotenuse = AC =q

So,

Now, LHS = q sin θ

= q = RHS

Hence Proved

Given: Sin θ

We know that,

Or

Let,

Perpendicular =AB =3k

and Hypotenuse =AC =5k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

In right angled ∆ ABC, we have

⇒ (AB)² + (BC)² = (AC)²

⇒ (3k)² + (BC)² = (5k)²

⇒ 9k² + (BC)² = 25k²

⇒ (BC)² = 25 k² –9k²

⇒ (BC)² = 16k²

⇒ BC =√16k²

⇒ BC =±4k

But side BC can’t be negative. So, BC = 4k

Now, we have to find the value of cos θ and tan θ

We know that,

The side adjacent to angle θ or base = BC =4k

Hypotenuse = AC =5k

So,

Now,

We know that,

Perpendicular = AB =3k

Base = BC =4k

So,

Now, LHS

= RHS

Hence Proved

Given:

To find: cos A sin B + sin A cos B

As, we have the value of sin A and cos B but we don’t have the value of cos A and sin B

So, First we find the value of cos A and sin B

We know that,

Or

Let,

Side opposite to angle A = 4k

and Hypotenuse = 5k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (P)² + (B)² = (H)²

⇒ (4k)² + (B)² = (5)²

⇒ 16 k² + (B)² = 25 k²

⇒ (B)² = 25 k² –16 k²

⇒ (B)² = 9 k²

⇒ B =√9 k²

⇒ B =±3k [taking positive square root since, side cannot be negative]

So, Base = 3k

Now, we have to find the value of cos A

We know that,

Side adjacent to angle A =3k

Hypotenuse =5k

So,

Now, we have to find the sin B

We know that,

Let,

Side adjacent to angle B =12k

Hypotenuse =13k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (B)² + (P)² = (H)²

⇒ (12k)² + (P)² = (13)²

⇒ 144 k² + (P)² = 169 k²

⇒ (P)² = 169 k² –144 k²

⇒ (P)² = 25 k²

⇒ P =√25 k²

⇒ P =±5k [taking positive square root since, side cannot be negative]

So, Perpendicular = 5k

Now, we have to find the value of sin B

We know that,

Now, cos A sin B + sin A cos B

Putting the values of sin A, sin B cos A and Cos B, we get

Given: tan A =√3 and

We know that,

Or

Let,

Side opposite to angle θ = 1k

and Hypotenuse = 2k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (AC)² + (BC)² = (AB)²

⇒ (1k)² + (BC)² = (2k)²

⇒ k² + (BC)² = 4k²

⇒ (BC)² = 4k² –k²

⇒ (BC)² = 3 k²

⇒ BC =√3k²

⇒ BC =k√3

So, BC = k√3

Now, we have to find the value of cos B

We know that,

The side adjacent to angle B = BC =k√3

Hypotenuse = AB =2k

So,

We know that,

Or

Given: tan A = √3

Let,

The side opposite to angle A =BC = √3k

The side adjacent to angle A =AB = 1k

where k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (1k)² + (√3k)² = (AC)²

⇒ (AC)² = 1 k² +3 k²

⇒ (AC)² = 4 k²

⇒ AC =√2 k²

⇒ AC =±2k

But side AC can’t be negative. So, AC = 2k

Now, we will find the sin A and cos A

Side opposite to angle A = BC = k√3

and Hypotenuse = AC = 2k

So,

Now, we know that,

The side adjacent to angle A = AB =1k

Hypotenuse = AC =2k

So,

Now, sin A. cos B – cos A. sin B

Putting the values of sin A, sin B cos A and Cos B, we get

Given:

Let,

Side opposite to angle A =BC = 1k

Side adjacent to angle A =AB = k√3

where, k is any positive integer

Firstly we have to find the value of BC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (√3k)² + (1k)² = (AC)²

⇒ (AC)² = 1 k² +3 k²

⇒ (AC)² = 4 k²

⇒ AC =√2 k²

⇒ AC =±2k

But side AC can’t be negative. So, AC = 2k

Now, we will find the sin A and cos A

Side opposite to angle A = BC = k

and Hypotenuse = AC = 2k

So,

Now, we know that,

Side adjacent to angle A = AB =k√3

Hypotenuse = AC =2k

So,

Now,

Given: tan B = √3

Let,

Side opposite to angle B =AC = √3k

Side adjacent to angle B =AB = 1k

where, k is any positive integer

Firstly we have to find the value of BC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (AC)² = (BC)²

⇒ (1k)² + (√3k)² = (BC)²

⇒ (BC)² = 1 k² +3 k²

⇒ (BC)² = 4 k²

⇒ BC =√2 k²

⇒ BC =±2k

But side BC can’t be negative. So, BC = 2k

Now, we will find the sin B and cos B

Side opposite to angle B = AC = k√3

and Hypotenuse = BC = 2k

So,

Now, we know that,

Side adjacent to angle B = AB =1k

Hypotenuse = BC =2k

So,

Now, sin A. cos B + cos A. sin B

Putting the values of sin A, sin B cos A and Cos B, we get

=1

Given

We know that,

Or

Let,

Side opposite to angle A = k

and Hypotenuse = k√2

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (P)² + (B)² = (H)²

⇒ (k)² + (B)² = (k√2)²

⇒ k² + (B)² = 2k²

⇒ (B)² = 2k² – k²

⇒ (B)² = k²

⇒ B =√k²

⇒ B =±k [taking positive square root since, side cannot be negative]

So, Base = k

Now, we have to find the value of tan A

We know that,

So,

Now, we have to find the tan B

We know that,

Let,

Side adjacent to angle B =k√3

Hypotenuse =2k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (B)² + (P)² = (H)²

⇒ (k√3)² + (P)² = (2k)²

⇒ 3k² + (P)² = 4k²

⇒ (P)² = 4k² –3 k²

⇒ (P)² = k²

⇒ P =√k²

⇒ P =±k [taking positive square root since, side cannot be negative]

So, Perpendicular = k

Now, we have to find the value of sin B

We know that,

So,

Now,

Now, multiply and divide by the conjugate of √3 – 1, we get

[∵ (a – b)(a+b) = (a² – b²)]

⇒ 2+√3

Given: tan A = 2 ⇒ tan²A = 4

We know that, sec² A = 1+ tan²A

⇒ sec² A = 1 + 4

⇒ sec² A = 5

⇒ sec A =√5

Now, we know that tan A

⇒ 2 =√5 sin A

Now, putting all the values in the given equation, we get

sec A. tan A+tan²A – cosec A

Given: cosec A =2

Now, we have to find

First, we simplify the above given trigonometry equation, we get

Taking the LCM, we get

[∵ cos²θ +sin² θ = 1]

[∵ cosec θ ]

⇒ cosec A

⇒ 2

Given:

We know that,

Or

Let,

Perpendicular =AB =k

and Hypotenuse =AC =2k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

In right angled ∆ ABC, we have

⇒ (AB)² + (BC)² = (AC)²

⇒ (k)² + (BC)² = (2k)²

⇒ k² + (BC)² = 4k²

⇒ (BC)² = 4k² –k²

⇒ (BC)² = 3k²

⇒ BC =√3k²

⇒ BC =k√3

So, BC = k√3

Now, we have to find the value of cos B

We know that,

Side adjacent to angle B or base = BC = k√3

Hypotenuse = AC =2k

So,

Now, LHS = 3 cos B – 4cos³ B

=RHS

Hence Proved

We know that,

Let AB =k√3 and BC = 2k

In right angled ∆ABC, we have

B² + P² = H²

⇒ (k√3)² + P² = (2k)²

⇒ P² + 3k² = 4k²

⇒ P² = 4k² – 3k²

⇒ P² = k²

⇒ P =√k²

⇒ P = ±k

⇒ P = k [taking positive square root since, side cannot be negative]

Now,

We know that,

Or

Now, LHS = 3sin θ – 4sin³ θ

⇒ 1 = RHS

Hence Proved

Given:

We know that,

Let,

BC = 4k and AC = 5k

where, k is any positive integer.

In right angled ∆ABC, we have

(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]

⇒ (AB)² + (4k)² = (5k )²

⇒ (AB)² + 16k² = 25k²

⇒ (AB)² = 25k² – 16k²

⇒ (AB)² = 9k²

⇒ AB = √9k²

⇒ AB =±3k [taking positive square root since, side cannot be negative]

Now, we have to find the value of other trigonometric ratios.

We, know that

Now, LHS

Now, RHS =

∴ LHS = RHS

Hence Proved

We know that,

Or

Let,

Side adjacent to angle B =AB = 12k

Side opposite to angle B =BC = 5k

where, k is any positive integer

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem

⇒ (AB)² + (BC)² = (AC)²

⇒ (12k)² + (5k)² = (AC)²

⇒ (AC)² = 144 k² +25 k²

⇒ (AC)² = 169 k²

⇒ AC =√169 k²

⇒ AC =±13k

But side AC can’t be negative. So, AC = 13k

Now, we will find the sin θ

Side opposite to angle B = BC = 5k

and Hypotenuse = AC = 13k

So,

Now, we know that,

Side adjacent to angle B = AB =12k

Hypotenuse = AC =13k

So,

Now, LHS = tan²B – sin² B

Now, RHS = sin⁴ B sec² B

Now, LHS = RHS

Hence Proved

Given:

Now, squaring both the sides, we get

⇒ p² = q² tan²θ …(1)

Now, solving LHS

Putting the value of p² in the above equation, we get

[∵ 1+ tan² θ = sec² θ]

(from Eq. (1))

[∵(a + b) (a – b) = (a² – b²)]

Now, we solve the RHS

[∵ 1+ tan² θ = sec² θ]

∴ LHS = RHS

Hence Proved

Given: BC =15cm and

We know that,

Or

Let,

Side opposite to angle B = 4k

and Hypotenuse = 5k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

⇒ (AC)² + (BC)² = (AB)²

⇒ (4k)² + (BC)² = (5)²

⇒ 16k² + (BC)² = 25k²

⇒ (BC)² = 25 k² –16 k²

⇒ (BC)² = 9 k²

⇒ BC =√9 k²

⇒ BC =±3k

But side BC can’t be negative. So, BC = 3k

Now, we have to find the value of cos B and tan B

We know that,

Side adjacent to angle B = BC =3k

Hypotenuse = AB =5k

So,

Now, tan B

We know that,

side opposite to angle B = AC =4k

Side adjacent to angle B = BC =3k

So,

Now,

= –1 = RHS

Hence Proved

First of all, we find the value of RS

In right angled ∆RQS, we have

(RQ)² + (QS)² = (RS)²

⇒ (8)² + (6)² = (RS)²

⇒ 64 + 36 = (RS)²

⇒ RS =√100

⇒ RS =±10 [taking positive square root, since side cannot be negative]

⇒ RS =10

Now, we find the value of QP

In right angled ∆RQP

(RQ)² + (QP)² = (RP)²

⇒ (8)² + (QP)² = (17)²

⇒ 64 + (QP)² = 289

⇒ (QP)² =289–64

⇒ (QP)² =225

⇒ QP =√225

⇒ QP =±15 [taking positive square root, since side cannot be negative]

⇒ QP =15

tan θ

Now, 3 tan θ – 2 sinα + 4cos α

⇒

⇒

⇒

Firstly, we find the value of AC

In right angled ∆ABC

(AB)² + (BC)² = (AC)²

⇒ (12)² + (5)² = (AC)²

⇒ 144+25 =(AC)²

⇒ (AC)² =169

⇒ AC =√169

⇒ AC =±13

⇒ AC =13 [taking positive square root since, side cannot be negative]

(i)

(ii)

Given: 5 sin² θ + cos² θ = 2

⇒ 5 sin² θ + (1– sin² θ) = 2 [∵ sin² θ + cos² θ = 1]

⇒ 5 sin² θ + 1 – sin² θ = 2

⇒ 4 sin² θ = 2 – 1

⇒ 4 sin² θ = 1

Given : 7 sin² θ + 3 cos² θ =4

⇒ 7 sin² θ + 3(1– sin² θ) = 4 [∵ sin² θ + cos² θ = 1]

⇒ 7 sin² θ + 3 –3 sin² θ = 4

⇒ 4 sin² θ = 4 – 3

⇒ 4 sin² θ = 1

Put the value of in given equation, we get

⇒

⇒

Now, we know that tan θ

Given : 4 cos θ+ 3 sin θ = 5

Squaring both the sides, we get

⇒ (4 cos θ+ 3 sin θ)² = 25

⇒ 16 cos² θ + 9 sin² θ + 2(4cos θ)(3sin θ)= 25 [∵ (a + b)² =a² +b² +2ab]

⇒ 16 cos² θ + 9 sin² θ + 24 cosθ sinθ = 25

Divide by cos² θ, we get

⇒ 16 + 9tan² θ + 24 tanθ = 25sec² θ

⇒ 16 + 9tan² θ + 24 tanθ = 25(1 + tan² θ) [∵ 1+ tan²θ = sec² θ]

⇒ 16 + 9tan² θ + 24 tanθ = 25+ 25 tan² θ

⇒ 16tan² θ – 24tanθ + 9 = 0

⇒ 16tan² θ – 12 tanθ – 12 tanθ +9 = 0

⇒ 4tanθ (4tan θ – 3) – 3(4tan θ – 3) = 0

⇒ (4tan θ – 3)² = 0

Given : 7 sin A + 24 cos A = 25

Squaring both the sides, we get

⇒ (7 sin A + 24 cos A)² = 625

⇒ 49 sin² A +576 cos² A + 2(7sin A) (24cos A) = 625 [∵ (a + b)² =a² +b² +2ab]

⇒ 49 sin² A +576 cos² A + 336 cosA sinA = 625

Divide by cos² θ, we get

⇒ 49tan² A +576+ 336 tanA = 625sec² A

⇒ 49tan² A +576+ 336 tanA = 625(1 + tan² A) [∵ 1+ tan²θ = sec² θ]

⇒ 49tan² A +576+ 336 tanθA = 625+625 tan² A

⇒ 576tan² A – 336tanA + 49 = 0

⇒ 576tan² A – 168 tanA – 168 tanA +49 = 0

⇒ 24tanθ (24tan A – 7) – 7(24tan A – 7) = 0

⇒ (24tan A – 7)² = 0

Given: 9 sin θ + 40 cos θ= 41

⇒ 9sinθ = 41 – 40 cosθ …(i)

Squaring both sides, we get

⇒ 81sin² θ = 1681+1600 cos² θ – 2(41) (40cos θ) [∵ (a – b)² =a² +b² –2ab]

⇒ 81 (1– cos² θ) =1681+1600 cos² θ – 3280cosθ

⇒ 81 – 81cos² θ = 1681 +1600cos² θ – 3280 cosθ

⇒ 1681cos² θ –3280cos θ +1600 = 0

⇒ (41)² cos² θ – 2(41) (40cos θ) + (40)² = 0

⇒ (41cos θ – 40 )² = 0

Now, putting the value of cos θ in Eq. (i), we get

tan A + sec A = 3

⇒ tanA = 3 – secA

Squaring both the sides, we get

⇒ tan² A =(3 – secA)²

⇒ tan² A = 9 + sec²A – 6sec A

⇒ sec² A – 1 = 9 + sec²A – 6sec A [∵ 1+ tan² A = sec² A]

⇒ –1 – 9 = –6secA

⇒ – 10 = –6sec A

Now, tan A + sec A = 3

⇒ sin A = 3cosA – 1

cosec A + cot A = 5

⇒ cotA = 5 – cosecA

Squaring both the sides, we get

⇒ cot² A =(5 – cosecA)²

⇒ cot² A = 25 + cosec²A – 10cosec A

⇒ cosec² A – 1 = 25 + cosec²A – 10cosec A [∵ 1+ cot² A = cosec² A]

⇒ –1 – 25 = –10cosecA

⇒ – 26 = –10cosec A

Now, cosec A + cot A = 5

tan θ+ sec θ = x

⇒ tan θ = x – sec θ

Squaring both sides, we get

⇒ tan² θ =(x – secθ)²

⇒ tan² θ = x² + sec²θ – 2xsec θ

⇒ sec² θ – 1 = x² + sec²θ – 2xsec θ [∵ 1+ tan² A = sec² A]

⇒ –1 – x² = –2xsecθ

Now,

tan θ = x – sec θ

= RHS

Hence Proved

Using the formula,

(a+b)² + (a – b)² = 2(a²+b²)

⇒ (cos θ +sin θ)² + (cos θ – sin θ)² = 2(cos²θ + sin² θ)

⇒ 1 + (cos θ – sin θ)² = 2(1)

⇒ (cos θ – sin θ)² = 2 –1

⇒ (cos θ – sin θ)² = 1

⇒ (cos θ – sin θ) =√1

⇒ (cos θ – sin θ) = ±1

(i) sin 30^o + cos 60^o

We know that,

>

So,
sin(30^o) + cos(60^o)

=1

(ii) sin² 45^o+cos²45^o

We know that,

So, sin² 45^o+cos²45^o

=1

(iii) sin 30^o + cos 60^o – tan45^o

So,
sin 30^o + cos 60^o – tan 45^o

=0

(iv)

We know that

tan(60^o) = √3

So,

=√4

= 2

(v) tan 60^o × cos30^o

tan(60^o) = √3

So,

tan 60^o × cos30^o

(i)

Given θ =45°

We know that,

tan(45^o) = 1

= 1+ 2

= 3

(ii) cos² θ – sin² θ

Given θ = 45°

So, cos² 45° – sin² 45°

= 0

We know that,

Now, putting the values

We know that,

tan (60^°) = √3

Now putting the values;

Multiplying and dividing by the conjugate of (1+√3)

[∵(a)² – (b)² = (a+b)(a-b)]

Multiplying and dividing by (-2)

= 3 - √3