Trigonometry

We have

Given: ∠ABC = 90°, AB = 8 cm, BC = 15 cm and CA = 17 cm

We know sin A is given by,

⇒ [∵, perpendicular is the side opposite to the angle A & hypotenuse is the side opposite to the right angle of that triangle]

⇒ …(i)

Also, cos A is given by

⇒

⇒ …(ii)

Now, tan A can be found out by two ways:

Method 1: tan A is given by,

⇒

⇒

Method 2: tan A can also be written as,

⇒ [from equations (i) & (ii)]

⇒

Thus, , and .

We have

We don’t need to find hypotenuse (PR) in the ∆PQR as tan θ = perpendicular/base.

⇒

And

⇒

tan P – tan R =

⇒ tan P – tan R =

⇒ tan P – tan R = 576/175

Thus, tan P – tan R = 576/175

We have

In ∆ABC, ∠ABC = 90° and ∠BAC = θ.

Using this information, we can say

AC = hypotenuse of the triangle

BC = perpendicular (side opposite to the angle θ or ∠BAC)

Using Pythagoras theorem,

(hypotenuse)² = (perpendicular)² + (base)²

⇒ (25)² = (24)² + (base)²

⇒ (AB)² = 625 – 576

⇒ (AB)² = 49

⇒ AB = √49 = 7 units

So, we have AB = 7 units, BC = 24 units and AC = 25 units.

Thus,

⇒

And

⇒

Thus, and .

Given that,

But

⇒

⇒ base = 12 and hypotenuse = 13

So, using Pythagoras theorem, we can say

(hypotenuse)² = (perpendicular)² + (base)²

⇒ (perpendicular)² = (hypotenuse)² – (base)²

⇒ (perpendicular)² = (13)² – (12)²

⇒ (perpendicular)² = 169 – 144 = 25

⇒ perpendicular = √25 = 5

Using perpendicular = 5, base = 12 and hypotenuse = 13, we can find out sin A and tan A.

Sin A is given by

⇒

And, tan A is given by

⇒

Thus, and .

Given that, 3 tan A = 4

⇒

But

⇒

⇒ perpendicular = 4 and base = 3

So, using Pythagoras theorem, we can say

(hypotenuse)² = (perpendicular)² + (base)²

⇒ (hypotenuse)² = (4)² + (3)²

⇒ (hypotenuse)² = 16 + 9 = 25

⇒ hypotenuse = √25 = 5

Using perpendicular = 4, base = 3 and hypotenuse = 5, we can find out sin A and cos A.

Sin A is given by

⇒

And, cos A is given by

⇒

Thus, and .

We have

In this ∆AOX,

cos A = cos X

and cos A is given by,

[∵, ]

Similarly,

⇒

⇒ AO = OX [clearly, since denominator from either sides cancel each other]

Now, if sides AO and OX of ∆AOX are equal.

Then, ∠A = ∠X [∵, In a triangle, angles opposite to the equal sides are also equal]

Hence, ∠A = ∠X

We have been given that,

And we have to solve for .

We know the formula: (x + y)(x – y) = x² – y²

Using this,

Also, we know the relationship between cos θ and sin θ which is given by

cos² θ + sin² θ = 1

⇒ cos² θ = 1 – sin² θ

So,

⇒

As we know,



So, B = 7 and P = 8

By Pythagoras theorem,

H² = P² + B²
H² = 8² + 7²
= 64 + 49
=113

H = √113

As we know,



So,

Given,

We know that, cosec²θ = 1 + cot²θ

We have

Given that, tan A = √3/1

And tan A is given by,

⇒

⇒ perpendicular = √3x and base = x

Then, we can use Pythagoras theorem in ∆ABC,

(hypotenuse)² = (perpendicular)² + (base)²

⇒ (hypotenuse)² = (√3x)² + (x)²

⇒ (hypotenuse)² = 3x² + x² = 4x²

⇒ hypotenuse = √(4x²) = 2x

We have, AB = √3x, BC = x and AC = 2x.

Using these values,

⇒

⇒

⇒ …(i)

Similarly,

⇒

⇒ …(ii)

Also,

⇒

⇒

⇒ …(iii)

Similarly,

⇒

⇒ …(iv)

We have to solve: sin A cos C + cos A sin C.

Substituting equations (i), (ii), (iii) & (iv) in above,

sin A cos C + cos A sin C =

⇒ sin A cos C + cos A sin C = 1/4 + 3/4

⇒ sin A cos C + cos A sin C = 4/4 = 1

Thus, sin A cos C + cos A sin C = 1.

To find: cos A cos C – sin A sin C.

From previous part of the question, we have

Using these values, we get

cos A cos C – sin A sin C =

⇒ cos A cos C – sin A sin C = √3/4 - √3/4 = 0

Thus, cos A cos C – sin A sin C = 0.

By trigonometric identities, we can say

sin 45° = 1/√2

and cos 45° = 1/√2

Adding them, we get

sin 45° + cos 45° = 1/√2 + 1/√2

⇒ sin 45° + cos 45° = 2/√2 = √2

Thus, sin 45° + cos 45° = √2.

Trigonometric identities:

cos 45° = 1/√2

sec 30° = 1/cos 30° = 2/√3 [∵, cos 30° = √3/2]

cosec 60° = 1/sin 60° = 2/√3 [∵, sin 60° = √3/2]

By trigonometric identities,

cot 45° = 1/tan 45° = 1/1 = 1

sec 30° = 1/cos 30° = 2/√3 [∵, cos 30° = √3/2]

cosec 60° = 1/sin 60° = 2/√3 [∵, sin 60° = √3/2]

sin 30° = 1/2

tan 45° = 1

cos 60° = 1/2

Putting all these values in , we get

Since, numerator is equal to denominator in the above calculation, we can say

By trigonometric identities,

sin 60° = √3/2

tan 45° = 1

cos 30° = √3/2

Putting these values in 2 tan² 45° + cos² 30° - sin² 60°, we get

2 tan² 45° + cos² 30° - sin² 60° = 2(1)² + (√3/2)² – (√3/2)²

⇒ 2 tan² 45° + cos² 30° - sin² 60° = 2

Thus, 2 tan² 45° + cos² 30° - sin² 60° = 2.

We have to solve:

Recall the trigonometric identities,

sin² θ + cos² θ = 1

& sec² α – tan² α = 1

Put θ = 30° and α = 60°, we get

sin² 30° + cos² 30° = 1

& sec² 60° - tan² 60° = 1

So,

Thus,

we know,

tan 30° = 1/√3

tan 45° = 1

Then, putting these values in the question, we get

⇒

And we know, tan 30° = 1/√3

But, sin 60° = √3/2

cos 60° = 1/2

& sin 30° = 1/2

Thus, option (C) is correct.

We know that,

tan 45° = 1

So, using this value of tangent, we can write

The answer has come out to be 0.

Thus, option (D) is correct.

We know that,

tan 30° = 1/√3

So, using this value of tangent, we can write

⇒

⇒

⇒

⇒

⇒

And we know, tan 60° = √3

But, cos 60° = 1/2

sin 60° = √3/2

& sin 30° = 1/2

Thus, option (C) is correct.

Let us first solve sin 60° cos 30° + sin 30° cos 60°.

We know,

sin 60° = √3/2

cos 30° = √3/2

sin 30° = 1/2

& cos 60° = 1/2

So, sin 60° cos 30° + sin 30° cos 60° =

⇒ sin 60° cos 30° + sin 30° cos 60° =

⇒ sin 60° cos 30° + sin 30° cos 60° = 1 …(i)

Now, for sin (60° + 30°):

sin (60° + 30°) = sin 90°

⇒ sin (60° + 30°) = 1 [∵, sin 90° = 1] …(ii)

By equations (i) & (ii), we can conclude that

sin (60° + 30°) = sin 60° cos 30° + sin 30° cos 60°

And infact in general, let 60° = x and 30° = y. Then,

sin (x + y) = sin x cos y + sin y cos x

Let us solve Left-Hand-Side:

cos (60° + 30°) = cos 90°

⇒ cos (60° + 30°) = 0 [∵, cos 90° = 0]

Now, solve for Right-Hand-Side:

⇒

⇒ cos 60° cos 30° - sin 60° sin 30° = 0

To find QR:

Since, tan θ = perpendicular/base

We know that,

⇒ [∵, PQ = 6 cm & tan 60° = √3]

⇒ QR = 6√3

Now, PR can be found by two ways -

1^st method: In ∆PQR, using Pythagoras theorem,

PR² = PQ² + QR² [∵, (hypotenuse)² = (perpendicular)² + (base)²]

⇒ PR² = 6² + (6√3)²

⇒ PR² = 36 + 108 = 144

⇒ PR = √144 = 12

2^nd Method:

Since, cos θ = base/hypotenuse

We know that,

⇒

⇒

⇒ PR = 2 × PQ

⇒ PR = 2 × 6 = 12

Thus, QR = 6√3 cm and PR = 12 cm.

We have

With given values, YZ = x and XZ = 2x, we can find out both angles.

For ∠YXZ:

Let ∠YXZ = θ, then

⇒

⇒

⇒ θ = sin^-1(1/2)

⇒ θ = 30° [∵, sin 30° = 1/2]

For ∠YZX = α, then

⇒

⇒

⇒ α = cos^-1(1/2)

⇒ α = 60° [∵, cos 60° = 1/2]

No, it is not correct to say that sin (A + B) = sin A + sin B.

Justification: Let’s justify it by showing contradiction.

Let it be true that, sin (A + B) = sin A + sin B.

Now, let A = 30° and B = 60°

Then,

sin (30° + 60°) = sin 30° + sin 60°

⇒ sin 90° = sin 30° + sin 60°

⇒ 1 = 1/2 + √3/2

⇒ 1 = (1 + √3)/2

But, it’s not true.

1 ≠ (1 + √3)/2

Hence, we have a contradiction.

And therefore, it’s not right to say that sin (A + B) = sin A + sin B.

By trigonometric identity, we have

tan (90° - θ) = cot θ

Replace θ = 54°

⇒ tan (90° - 54°) = cot 54°

⇒ tan 36° = cot 54° [∵, 90° - 54° = 36°]

Now,

⇒

By trigonometric identity, we can say

cos (90° - θ) = sin θ

Now, just replace θ by 78°.

We get

cos (90° - 78°) = sin 78°

⇒ cos 12° = sin 78° [∵, 90° - 78° = 12°]

Now, cos 12° - sin 78° = sin 78° - sin 78°

⇒ cos 12° - sin 78° = 0

By trigonometric identity, we can say

cosec (90° - θ) = sec θ

Replace θ by 59°.

We get

cosec (90° - 59°) = sec 59°

⇒ cosec 31° = sec 59°

Now, cosec 31° - sec 59° = sec 59° - sec 59°

⇒ cosec 31° - sec 59° = 0

By trigonometric identities, we can say

sin (90° - θ) = cos θ

And sec θ = 1/cos θ

Replace θ by 75°.

We get

sin (90° - 75°) = cos 75°

⇒ sin 15° = cos 75° [∵, 90° - 75° = 15°]

And sec 75° = 1/cos 75°

Using these values, we can solve the given expression.

sin 15° sec 75° = sin 15°/cos 75°

⇒ sin 15° sec 75° = cos 75°/cos 75°

⇒ sin 15° sec 75° = 1

By trigonometric identities, we can say

tan (90° - θ) = cot θ

And tan θ = 1/cot θ

Replace θ by 64°.

We get

tan (90° - 64°) = cot 64°

⇒ tan 26° = cot 64° [∵, 90° - 64° = 26°]

And tan 64° = 1/cot 64°

Using these values, we can solve for the given expression.

tan 26° tan 64° = cot 64° tan 64°

⇒ tan 26° tan 64° = cot 64°/cot 64°

⇒ tan 26° tan 64° = 1

We have

LHS = tan 48° tan 16° tan 42° tan 74° = (tan 48° tan 42°)(tan 16° tan 74°)

We know the trigonometric identities, we can say

tan (90° - θ) = cot θ

And tan θ = 1/cot θ

First, replace θ by 42°.

tan (90° - 42°) = cot 42°

⇒ tan 48° = cot 42° …(1)

And tan 42° = 1/cot 42° …(2)

Now, replace θ by 74°.

tan (90° - 74°) = cot 74°

⇒ tan 16° = cot 74° …(3)

And tan 74° = 1/cot 74° …(4)

Using equations (1), (2), (3) & (4), we get

LHS = tan 48° tan 16° tan 42° tan 74° = (tan 48° tan 42°)(tan 16° tan 74°)

= (cot 42°/cot 42°)(cot 74°/cot 74°)

= 1 = RHS

Thus, tan 48° tan 16° tan 42° tan 74° = 1.

We have

LHS = cos 36° cos 54° - sin 36° sin 54°

We know the trigonometric identity,

cos (90° - θ) = sin θ

First, replace θ by 54°.

We get, cos (90° - 54°) = sin 54°

⇒ cos 36° = sin 54° …(1)

Now, replace θ by 36°.

We get, cos (90° - 36°) = sin 36°

⇒ cos 54° = sin 36° …(2)

Using equations (1) & (2), we get

LHS = cos 36° cos 54° - sin 36° sin 54° = sin 54° sin 36° - sin 54° sin 36°

= 0 = RHS

Thus, cos 36° cos 54° - sin 36° sin 54° = 0.

Given that, 2A is an acute angle.

⇒ 2A < 90°

So, using trigonometric identity, we can say that

cot (90° - 2A) = tan 2A [∵, cot (90° - θ) = tan θ]

Now, replace tan 2A by cot (90° - 2A) in the given question.

tan 2A = cot (A – 18°)

⇒ cot (90° - 2A) = cot (A – 18°)

Now, we can compare the degrees from above, we get

90° - 2A = A – 18°

⇒ 2A + A = 90° + 18°

⇒ 3A = 108°

⇒ A = 108°/3

⇒ A = 36°

Thus, the value of A is 36°.

Given that, A and B are acute angles.

⇒ A < 90° & B < 90°

So, using trigonometric identity, we can say

tan (90° - B) = cot B [∵, tan (90° - θ) = cot θ]

Replace cot B of RHS by tan (90° - B) in the given question.

tan A = cot B

⇒ tan A = tan (90° - B)

Now, comparing the degrees from the above, we get

A = 90° - B

⇒ A + B = 90°

Hence, proved that A + B = 90°.

If A, B and C are interior angles of ∆ABC, then we can say that

A + B + C = 180° (by angle sum property of a triangle)

⇒ A + B = 180° - C

Take LHS:

⇒

⇒

[∵, tan (90° - θ) = cot θ, where θ = C/2 here]

Hence, .

Given: sin 75° + cos 65°.

We can write,

75° = 90° - 15°

& 65° = 90° - 25°

Then, sin 75° + cos 65° = sin (90° - 15°) + cos (90° - 25°)

⇒ sin 75° + cos 65° = cos 15° + sin 25°

[∵, sin (90° - θ ) = cos θ & cos (90° - θ) = sin θ]

In cos 15° + sin 25°, 15° & 25° both are angles between 0° and 45°.

Thus, answer is sin 75° + cos 65° = cos 15° + sin 25°.

We have

⇒

⇒

⇒ [∵, (a + b)(a – b) = a² – b²]

⇒ [∵, (a + b)² = a² + b² + 2ab]

⇒ [∵, sin² θ + cos² θ = 1]

⇒

Thus, (1 + tan θ + sec θ)(1 + cot θ – cosec θ) = 2.

We have

(sin θ + cos θ)² + (sin θ – cos θ)² = ((sin θ + cos θ) + (sin θ – cos θ))² – 2(sin θ + cos θ)(sin θ – cos θ) [∵, a² + b² = (a + b)² – 2ab]

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = (sin θ + cos θ + sin θ – cos θ)² – 2(sin² θ – cos² θ)

[∵, (a + b)(a – b) = a² – b²]

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = (2 sin θ)² – 2 sin² θ + 2 cos² θ

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = 4 sin² θ – 2 sin² θ + 2 cos² θ

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

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = 2 (sin² θ + cos² θ)

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = 2 [∵, sin² θ + cos² θ = 1]

Thus, (sin θ + cos θ)² + (sin θ – cos θ)² = 2.

We have

⇒

⇒ [∵, (1 – cos² θ) = sin² θ & (1 – sin² θ = cos² θ]

⇒ (sec² θ – 1)(cosec² θ – 1) = 1

Thus, (sec² θ – 1)(cosec² θ – 1) = 1.

Using trigonometric identities,

cosec θ = 1/sin θ & cot θ = cos θ/sin θ

LHS = (cosec θ – cot θ)²

As we know, sin² θ = 1 – cos² θ

And (1 – cos² θ) = (1 + cos θ)(1 – cos θ) [by (a² – b²) = (a + b)(a – b)]

⇒ sin² θ = (1 + cos θ)(1 – cos θ) …(i)

So, using equation (i),

LHS = (cosec θ – cot θ)² =

= RHS

Hence, we have got

(cosec θ – cot θ)² = .

Using trigonometric identity,

sin² A + cos² A = 1

⇒ cos² A = 1 – sin² A

Take Left hand side:

LHS =

= sec A + tan A = RHS

Thus, .

Using trigonometric identity, we have

cot A = 1/tan A

Take left hand side,

LHS =

= tan² A = RHS

Thus, = tan² A.

Take left hand side of the given question:

LHS = 1/cos θ – cos θ

= (1 – cos² θ)/cos θ

= sin² θ/cos θ [∵, sin² θ + cos² θ = 1 ⇒ sin² θ = 1 – cos² θ]

= sin θ × sin θ/cos θ

= sin θ × tan θ [∵, tan θ = sin θ/cos θ]

= tan θ sin θ = RHS

Thus, .

By trigonometric identities, sec A = 1/cos A & tan A = sin A/cos A

Using these identities, we have

sec A (1 – sin A)(sec A + tan A) =

⇒ sec A (1 – sin A)(sec A + tan A) =

⇒ sec A (1 – sin A)(sec A + tan A) =

⇒ sec A (1 – sin A)(sec A + tan A) =

⇒ sec A (1 – sin A)(sec A + tan A) = [∵, sin² A + cos² A = 1 ⇒ cos² A = 1 – sin² A]

⇒ sec A (1 – sin A)(sec A + tan A) = 1

Thus, sec A (1 – sin A)(sec A + tan A) = 1.

Take left hand side of the given equation:

LHS = (sin A + cosec A)² + (cos A + sec A)²

= sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A

= (sin² A + cos² A) + 2 sin A cosec A + 2 cos A sec A + cosec² A + sec² A

we know that,

=

=

= 5 + 1/(sin² A cos² A) …(i)

Now, take right hand side of the equation:

RHS = 7 + tan² A + cot² A

=

=

=

=

= 5 + 1/(sin² A cos² A) …(ii)

From equation (i) & (ii),

LHS = RHS

Hence, proved.

We have

(1 – cos θ)(1 + cos θ)(1 + cot² θ) = [(1 – cos θ)(1 + cos θ)](1 + cot² θ)

⇒ (1 – cos θ)(1 + cos θ)(1 + cot² θ) = (1 – cos² θ)(1 + cot² θ) [∵, (a + b)(a – b) = a² – b²]

⇒ (1 – cos θ)(1 + cos θ)(1 + cot² θ) = sin² θ × (1 + cos² θ/sin² θ) [∵, (1 – cos² θ) = sin² θ & cot² θ = cos² θ/sin² θ]

⇒ (1 – cos θ)(1 + cos θ)(1 + cot² θ) = sin² θ × (sin² θ + cos² θ)/sin² θ

⇒ (1 – cos θ)(1 + cos θ)(1 + cot² θ) = sin² θ + cos² θ

⇒ (1 – cos θ)(1 + cos θ)(1 + cot² θ) = 1 [∵, sin² θ + cos² θ = 1]

Thus, (1 – cos θ)(1 + cos θ)(1 + cot² θ) = 1.

Given that, sec θ + tan θ = p.

By trigonometric identity, we have

sec² θ – tan² θ = 1

So, sec² θ – tan² θ = 1

⇒ (sec θ – tan θ) (sec θ + tan θ) = 1

⇒ sec θ – tan θ = 1/(sec θ + tan θ)

⇒ sec θ – tan θ = 1/p [given]

Hence, sec θ – tan θ = 1/p.

Given that, cosec θ + cot θ = k

⇒ [∵, cosec θ = 1/sin θ & cot θ = cos θ/sin θ]

⇒ (1 + cos θ)/sin θ = k

⇒ 1 + cos θ = k sin θ

Squaring both sides, we get

(1 + cos θ)² = (k sin θ)²

⇒ (1 + cos θ)² = k² sin² θ

⇒ (1 + cos θ)² = k² (1 – cos² θ) [∵, sin² θ + cos² θ = 1 ⇒ sin² θ = 1 – cos² θ]

⇒ (1 + cos θ)² = k² (1 – cos θ) (1 + cos θ) [∵, a² – b² = (a + b) (a – b)]

⇒ 1 + cos θ = k² (1 – cos θ)

⇒ 1 + cos θ = k² – k² cos θ

⇒ k² cos θ + cos θ = k² – 1

⇒ cos θ (k² + 1) = k² – 1

⇒ cos θ = (k² – 1)/(k² + 1)

Thus, cos θ = (k² – 1)/(k² + 1).

Hence, proved.