Trigonometric Ratios

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

In a , right angled at B, AB = 24 cm, BC = 7 cm. Therefore, By Pythagoras Theorem,

Therefore,

By Pythagoras theorem we know that,

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

we also know that,





Note: When finding any trigonometric ratio, the main part is to decide perpendicular and base for that angle. Perpendicular is the side opposite to the angle for which we are calculating. For example from above figure if we are calculating sin R, then side opposite to R is PQ, So PQ will be perpendicular, PR is Hypotenuse and the side left out will be base.

Given

By Pythagoras theorem,

Given: In , right angled at Q, PQ = 4 cm and RQ = 3 cm.

To find: the values of sin P, sin R, sec P and sec R.

Solution:

In triangle PQR, , PQ = 4 cm and RQ = 3 cm

By Pythagoras theorem,

Use the formula sinθ = perpendicular / hypotenuse

Secθ = hypotenuse / base

NOTE: Always check on which angle you are asked to find any trignometric
value and take perpendicular,base,hypotenuse accordingly.

(i)

(ii)

Thus, it is true.

Given:

To find: the value of

Solution:


Take cosΘ common,

And use the formula:



Solve,

Thus, it is true.

Given:3 tanθ = 4
To find: the value of
Solution:

Here,3 tanθ = 4

Use the formula,

Solve,

Given:
To prove: ...... (1)
Solution:

Consider LHS of eq. (1)
Take b cosθ common from both numerator and denominator.

Solve using the formula:


Put the value of tanθ to get,



hence proved

Given:
To prove: sin (1-tan) =

Proof:
we know,

Where B is base and H is hypotenuse of the right angled triangle.
We construct a right triangle ABC right angled at B such that
Perpendicular is AB, Base is BC = 12 and hypotenuse is AC = 13.
In the triangle ABC,
By Pythagoras theorem, we have

13² = AB²+12²
169 = AB²+144
169-144=AB²
25=AB²


So,

Put the values in sin (1-tan) to find its value,

_{Hence Proved.}

_{Hence Proved.}

_{Hence Proved.}

Given:


To find: the value of


Solution:


Since




So implies:

Perpendicular = AC = 12, Hypotenuse = BC = 13

Draw a right angled triangle at C,




By Pythagoras theorem,

AB² = AC² + BC²

⇒ (13)² = (12)² + BC²

⇒ BC² = (13)² - (12)²

⇒ BC² = 169 - 144

⇒ BC² = 25



⇒ BC = 5

Since cosθ = Base/Hypotenuse and tanθ = Perpendicular/Base

⇒

Substitute the known values in ,

Given:

To find: the value of

Solution:

We know:




By applying Pythagoras theorem,we have


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

⇒ BC² = AB² + AC²

⇒ BC² = 3² + 4²

⇒ BC² = 9 + 16

⇒ BC² = 25

⇒ BC =

⇒ BC = 5



Use:









Substitute the known values,

Given:

To find: in terms of a and b.
Solution:

We will construct a right angled triangle, right angled at B such that,

Perpendicular=BC=a and hypotenuse=AC=b
AC² =AB² + BC²
b² =AB² + a²
AB² = b² - a<sup>2


Use the formula:</sup>

Solve,

Given:


To show:

Solution:

Since tanθ = perpendicular / base

So we construct a right triangle ABC right angled at C such that

∠ABC=θ and AC = Perpendicular = 20
BC = base = 21

By Pythagoras theorem,

AB² = AC² + BC²

⇒ AB² = (20)² + (21)²

⇒ AB² = 400 + 441

⇒ AB² = 841

⇒ AB = √841

⇒ AB = 29


As sinθ = perpendicular / hypotenuse

cosθ = base / hypotenuse

So,


Hence proved

In a right angled triangle ABC,

We have, opposite sides of equal angles are equal. Therefore, In a right angled triangle ABC

In a right angled triangle APQ,

In a ΔABC, right angled at A,

tan C = √3

i.e ∠C = 60° and ∠B = 90 - 60 = 30°

Sin C = Sin 60° = √3/2

Cos C = Cos 60° = 1/2

Sin B = Sin 30° = 1/2

Cos B = Cos 30° = √3/2

According to the question,

sin B cos C + cos B sin C

= (1/2) (1/2) + (√3/2) (√3/2)

= 1/4 + 3/4

= 1

(i) The value of tan 90° is greater than 1. Therefore, given statement is false.

(ii) as 12 is the hypotenuse largest side. Therefore, given statement is true.

(iii) cos A is the abbreviation used for cosine of angle A Therefore, given statement is true.

(iv) cot A is not the product of cot and A. Therefore, given statement is false.

(v) Since, the hypotenuse is the longest side whereas in sin A = , 3 which is the denominator and cannot be hypotenous.

Given:
To find: The value of above equation.
Solution:
Use the values:

Solve,

Given :
To find : The value of
Solution :
Use the values:

Solve,

Hence the value of is 1.

Given: 4 (sin⁴ 60° + cos⁴ 30°)-3 (tan² 60° - tan² 45°) + 5cos² 45°

To find: The value of 4 (sin⁴ 60° + cos⁴ 30°)-3 (tan² 60° - tan² 45°) + 5cos² 45°.

Solution:

We know,



Substitute the above values in 4 (sin⁴ 60° + cos⁴ 30°)-3 (tan² 60° - tan² 45°) + 5cos² 45°,

Solve,

Given: ( cos 0^°+sin 45^°+sin 30^° ) ( sin 90^° - cos 45^° + cos 60^° )
To find: The value of the above.
Solution:
Use the formulas:


Solve,
( cos 0^°+sin 45^°+sin 30^° ) ( sin 90^° - cos 45^° + cos 60^° )




Use the identity: (a-b)(a+b)=a²-b²

Given:
To find: The value of
Solution:
Apply cross multiplication in the given expression,

Given :
To find : The value of x.
Solution :
Put the values:

solve,

_(i)

(ii)


Use the formula:

Hence proved

(iii)


Use the formula,

Hence proved

(iv)


Use the formula,

Hence proved

(i)

(ii)

(iii)

(i)

Use the formula,

Hence proved

(ii)


Use the formula,

For finding the values of sin 15º and cos 15°, we can split 15° into two angles such that,
15° = 45° - 30°

or we also can split 15° as 15° = 60° - 45°
You can use either way, answer won't change.
Formula to use for calculating this value is already given in the question,
sin(A - B) = sin A cos B - cos A sin B (1)
cos(A - B) = cos A cos B + sin A sin B (2)
Put A = 60° and B = 45°
A - B = 15°
Now put the values in formula 1 and 2,

And,

Thus we have,

In a right triangle ABC, right angled at C, if = 60° and AB = 15 units. Therefore,

And,

Ifis a right triangle such that = 90°, = 45° and BC = 7 units. Therefore,

And,

If rectangle ABCD AB = 20 cm, = 60°. Therefore,

And,

_{Therefore, AC = BD = 40 cm}

On solving both equations, we get

A = 45° and B = 15°

On solving both equations, we get

A = 45°, B = 15°

In a right angled at B, _{, therefore,}

_(i)

(ii)

Given: sin (A + 2B) = and cos (A + 4B) = 0
To find: The values of acute angles A and B.
Solution:
We know,

So,

Solve eq. (1) and eq. (2) to get the values of a and b.
Subtract eq. (1) from eq. (2),
⇒A+4B-(A+2B)=90°-60°

Hence the values of A and B are A = 30°, B = 15°

Given: In ΔPQR, right-angled at Q, PQ = 3 cm and PR = 6 cm.

To find:

Solution:

⇒ R = 30^°

Now, in ΔPQR

As sum of all angles of a triangle is 180^°.

∠P = 180^° - ( 90^° + 30^° )

∠P = 180^° - 120^°

∠P = 60^°

Use:



(i)

(ii)

(iii)

(iv)

(v)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

Use the formula:






(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Given :
To find : Expression in terms of angles between 0° and 30°.
Solution :
Use the values:

Solve,

(i) Since, A, B, C, are the interior angles of a triangle ABC.

Therefore,

Hence proved.

(ii)

Hence proved.

(i)

Which is equal to RHS.

Hence Proved

(ii)

_Proved

(iii)

_Proved

(iv)

_Proved

In the given parts use:

(i)

Which is equal to RHS.

(ii)

(iii)

(iv)

(v)
solve LHS,

(i)

(ii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(i) Since, A, B, C are the interior angles of a

(ii)

Given:, where and are acute angles.
To find: The value of .
Solution:
since,

So,

With the help of Minimum-Maximum Value Table we can find the Value of sin θ

Therefore,

Minimum Value of sin θ = - 1 and

Maximum Value of sin θ = 1

Given,

cos θ = 3/5

In ∆ ABC,

AC² = AB² + BC²

(5)² = AB² + (3)²

25 = AB² + 9

25 – 9 = AB²

16 = AB²

4 = AB

Therefore,

AB = 4

As,

sin θ = AB/AC = 4/5

tan θ = AB/BC = 4/3

Now,

With the help of Minimum-Maximum Value Table we can find the Value of cos θ

Therefore,

Minimum Value of cos θ = - 1 and

Maximum Value of cos θ = 1

Given,

tan θ = a/b

In ∆ ABC,

AC² = AB² + BC²

AC² = a² + b²

Therefore,

Now,

As we know,

And,

Maximum value of cos θ = 1

So,

The maximum value of 1/sec θ is 1

Given,

5 tan θ – 4 = 0

So,

5 tan θ = 4

tan θ = 4/5

In ∆ ABC,

AC² = AB² + BC²

AC² = (4)² + (5)²

AC² = 16 + 25

AC² = 41

AC = √41

Therefore,

sin θ = AB/AC = 4/√41

cos θ = BC/AC = 5/√41

Now,

As we know,

And,

Maximum value of sin θ = 1

So,

The maximum value of 1/cosec θ is 1.

Given:16 cot x = 12

To find: The value of .

Solution:

cot x = 12/16

cot x = 3/4

Also cotθ = base/perpendicular

So we now construct a right triangle ABC, right angled at B such that

∠BAC = θ, Base = 3 and perpendicular = 4

In ∆ ABC,

AC² = AB² + BC²

AC² = (3)² + (4)²

AC² = 9 + 16 = 25

AC = 5

Therefore,

sin x = AB/AC = 4/5

cos x = BC/AC = 3/5

Now,

Given,

8 tan x = 15

So,

tan x = 15/8

In ∆ ABC,

AC² = AB² + BC²

AC² = (15)² + (8)²

AC² = 225 + 64 = 289

AC = 17

Therefore,

sin x = AB/AC = 15/17

cos x = BC/AC = 8/17

Now,

Given,

tan θ = 4/5

As we know,

And,

Now,

Therefore,

Given,

tan θ = 1/√7

In ∆ ABC,

AC² = AB² + BC²

AC² = (1)² + (√7)²

AC² = 8

AC = 2√2

Therefore,

Now,

Given,

cos θ = 2/3

We know,

So,

Therefore,

Given,

tan θ = 3/4

In ∆ ABC,

AC² = AB² + BC²

AC² = (3)² + (4)²

AC² = 9 + 16 = 25

AC = 5

Therefore,

sin θ = p/h = 3/5 and

cos θ = b/h = 4/5

Now putting these values in the given equation we get,

cos² θ – sin² θ =

Given,

3 cot θ = 4

So,

cot θ = 4/3

As we know,

And,

Now,

Given,

tan² θ = 8/7

Now,

Given,

tan θ = 1/√5

In ∆ ABC,

AC² = AB² + BC²

AC² = 1² + (√5)²

AC² = 1 + 5 = 6

AC = √6

We have,

cosec θ = AC/AB = √6/1

sec θ = AC/BC = √6/√5

Now,

Given,

3 cos θ = 5 sin θ

In ∆ABC,

AC² = AB² + BC²

AC² = (3)² + (5)²

AC² = 9 + 25 = 34

AC = √34

Therefore,

sin θ = 3/√34

cos θ = 5/√34

sec θ = √34/5

Now,

Given,

cot θ = 1/√3

In ∆ABC,

AC² = AB² + BC²

AC² = (√3)² + 1²

AC² = 3 + 1 = 4

AC= 2

We get;

cos θ = BC/AC = 1/2

sin θ = AB/AC = √3/2

Now,

Given,

tan² 45° - cos² 30° = x sin 45° cos 45° …….(i)

put the values in equation (i),

Given,

tan A = 3/4 and

A + b = 90°

So we have,

A = 90° - B

So,

tan A = tan (90° - B) = 3/4

∵ cot B = tan A

Therefore,

cot B = 3/4

cos² 17° - sin² 73° = cos² 17° - sin²(90° -17°)

= cos²17° - cos² 17° = 0

Given,

cos B = 3/5 and

A + B = 90°

So we have,

B = 90° - A

So,

cos B = cos (90° - A) = 3/5

∵ cos B = sin A

Therefore,

sin A = 3/5

Given,

√3 sin θ = cos θ

sin θ / cos θ = 1/√3
tan θ = 1/ √3 …(the value of tan 30° is 1/√3)

Therefore,
tan θ = tan 30°
θ = 30°

Given,

8x = 8

x = 8/8 = 1

So, the value of x = 1

cos 1° cos 2° cos 3°…. cos 179° cos 180°
= cos 1° × cos 2° × cos 3° × .....× cos 90° × ..... × cos 179°

As we know,

cos 90° = 0

So,
= cos 1° × cos 2° × cos 3° × .....× 0 × ..... × cos 179°
= 0

As, cos 90° has the value 0 that’s why the whole answer will be zero.

Given,

A + B = 90°

B = 90° - A

sin B = sin (90° - A)

sin B = cos A

Taking the reciprocal,

cosec B = sec A

Or

sec A = cosec B

tan 10° × tan 15° × tan 75° × tan 80°

= tan 10 × tan 15 × tan(90-75) × tan (90-80)
= tan10 × tan15 × cot 15 × cot10
As we know,

tan θ × cot θ = 1
So,
tan 10 × cot 10 × tan 15 × cot 15
= 1

Given: x sin (90° - 0) cot (90° - θ) = cos (90° - θ)

To find: The value of x.

x sin (90° - 0) cot (90° - θ) = cos (90° - θ)

⇒ x cos θ.tan θ = sin θ

⇒

⇒

⇒ x = 1

Hence, the value is x = 1

Given,

tan A = 3/4

A + B = 90°
B = 90 – A

So,
cot B = cot (90-A)
cot B = tan A

Therefore,
cot B = 3/4

Given,

tan A = 5/12

We get,

Now,

(sin A + cos A) sec A = (sin A + cos A)× 1/cos A

= tan A + 1

Therefore,

cot A = 17/12

Given,

x tan 45° cos 60° = sin 60° cot 60°

(x)×(1)× 1/2 = √3/2 × 1/√3

x/2 = 1/2

x = 2/2 = 1

So the value of x is 1.

Let suppose A, B and C are the angles of a triangle ABC,

In ∆ABC,

∠ A = (a – d)

∠ B = a

∠ C = a + d

Now, form an increasing A.P

As we know Sum of all the angle of a triangle is 180°,

Therefore,

∠A + ∠B + ∠C = 180°

(a – d) + a + (a + d) = 180°

3a = 180°

a = 180/3 = 60°

From the table,

sin B = sin A = sin 60°

= √3/2

Given,

Sec²θ = 3

So,

Sec θ = √3 = h/b = k

Therefore,

h = √3k, b = k

In ∆ABC,

h² = p² + b²

(√3k)² = p² + (k)²

3k² = p² + k²

3k² – k² = p²

2k² = p²

√2 K = p

We know,

tan θ = p/b = √2k/k = √2

cosec θ = h/p = √3k/√2k = √3/√2

Put these value in,

tan 1° × tan 2° × tan 3° × ………. × tan 89°

As we know tan (90 – θ) = cot θ

So here we get,

tan (90 – 89) × tan (90 - 88) × tan (90 - 87) × ……….× tan 87 × tan 88 × tan 89°

cot 89° × cot 88° × cot 87°…………….. tan 45°× tan 46° …………….× tan 87°× tan 88°× tan 89°

∵ cot θ = 1/tan θ

Therefore,

(cot 89° × tan89°)(cot 88° × tan88°)(cot 87° × tan87°)…… (cot 46° × tan46°)(tan 45°)

= tan 45°

= 1

Given,

cos 1° cos 2° cos 3° ………………..cos 180°

cos 1° × cos 2° × cos 3° …… cos 89° × cos 90° × cos 91° ……….cos 180°

As we know from the table,

cos 90° = 0

Therefore,

cos 1° × cos 2° × cos 3° …… cos 89° × 0 × cos 91° ……….cos 180°

= 0

Given,

tan 10° tan 15° tan 75^o tan 80°

As we know,

tan (90 – θ) = than θ

Therefore,

⇒ tan (90°-80°) tan (90°-75°) tan 75^o tan 80°

⇒ cot 80° cot 75° tan 75° tan 80°

⇒ (cot 80° tan 80°) (cot 75° tan 75°)……… [ cot θ = 1/tanθ ]

⇒

⇒ (1)(1) = 1

Given,

…….. (i)

∵ cos (90 – θ) = sin θ cos (90 – θ) = cos θ

sec (90 – θ) = cosec θ cot (90 – θ) = tan θ

cosec (90 – θ) = sec θ tan (90 – θ) = cot θ

Putting these values in (i),

We get,

∵ cosec θ = 1/sin θ and

sec θ = 1/cos θ

⇒

⇒ 1 + 1 = 2

Given,

θ and 2θ – 45° are acute angle,

sin θ = cos (2θ – 45)………….. (i)

[∵ cos (90 – θ) = sin θ]

Putting these value in equation (i),

Cos (90 – θ) = cos (2θ – 45°)

90 - θ = 2 θ – 45°

90 + 45 = 3 θ

3 θ = 135

θ = 135/3 = 45

tan θ = tan 45° = 1

Given,

5θ and 4 θ are acute angles,

Therefore,

5θ + 4θ = 90°

9θ = 90°

θ = 90/9 = 10°

Then value of-

2 sin 3θ - √3 tan 3 θ

Putting value of θ = 10°,

We get,

⇒ 2 sin 3(10) - √3 tan 3(10)

⇒ 2 sin 30° - √3 tan 30° [∵ sin 30° = 1/2 and tan 30° = 1/√3]

⇒ 2× 1/2 - √3 x

⇒ 1 – 1 = 0

Given,

A + B = 90°

B = 90° - A

Putting this Value in the given equation we get,

∵ tan (90 – A) = Cot A sin(90 – A) = cos A

cot (90 – A) = tan A

sec (90 – A) = cosec A

[∵ cot A = 1/tan A and cosec A = 1/sin A]

⇒ 1 + tan² A – 1

∵ A + B = 90°

A = 90 – B

So,

⇒ tan²(90 – B)

⇒ cot²B

Given,

[∵ sin 60° = √3/2]

Given,

∵ tan 45 = 1

Put this value,

We get;

sin 0°

Since sin 0° = 0

Sin 2 A = 2 sin A

[∵ 2A = 2 sin A. Cos A]

⇒ 2 sin A . cos A = 2 sin A

⇒ cos A = 1 = cos 0°

⇒ A = 0°

tan 60° [∵ tan 60° = √3]

Given,

A, B and C are the interior angles of ∆ ABC,

Therefore,

A + B + C = 180°

B + C = 180° - A

Now put this value in the given equation we get,

[∵ sin (90 – θ) = cos θ]

Given,

cos θ = 2/3 = b/h = k

2sec² θ + 2 tan²θ – 7

b = 2k, h = 3k

In ∆ABC,

h² = p² + b²

⇒ (3k)² = p² + (2k)²

⇒ 9k² = p² + 4k²

⇒ p² = 9k² – 4k²

⇒ p² = 5k²

⇒ p = √5k

Then,

Sec θ = h/b = 3k/2k = 3/2 and

Tan θ = p/b = √5k/2k = √5/2

⇒ 2 sec² θ + 2 tan² θ – 7

⇒

⇒

⇒

⇒

⇒ tan 5° x tan 30° x 4 tan 85°

As,

tan (90 – θ) = cot θ

Therefore,

⇒ tan (90 – 85) x tan 30° x 4 tan 85°

⇒ cot 85° × tan 85° × 4 × tan 30°

⇒ 1/tan 85 × tan 85° × 4 × tan 30°

As we know,

tan 30° = 1/√3

⇒ 4 × tan 30°

⇒ 4 × (1/√3) = 4/√3

Given,

As we know,

Cot 90° = 0

Therefore,

1 + 0 = 1

As we know that sum of the angles of the straight line is 180°,

Therefore,

∠ θ + ∠90 + ∠ϕ = 180

∠ θ + ∠ϕ = 90

In ∆ABC,

sin θ = 4/5 = p/h

Putting θ = 90 - ϕ

We get,

sin (90 - ϕ) = 4/5

As,

Sin (90 - ϕ) = cos ϕ

cos ϕ = 4/5

Given,

AD = 4cm

BD = 3 cm

CB = 12 cm

In ∆ABC,

AB² = AD² + BD²

AB² = 4² + 3²

AB = √16 + 9 = 5 cm

Then,

cot θ = CB/AB = 12/5