Chapter 8: Introduction to Trigonometry & Its Equations

(B) 3/4

According to the question,

cos A = 4/5 …(1)

We know,

tan A = sinA/cosA

To find the value of sin A,

We have the equation,

sin² θ +cos² θ =1

So, sin θ = √ (1-cos² θ)

Then,

sin A = √ (1-cos² A) …(2)

sin² A = 1-cos² A

sin A = √(1-cos² A)

Substituting equation (1) in (2),

We get,

Sin A = √(1-(4/5)²)

= √(1-(16/25))

= √(9/25)

= ¾

Therefore,

(A) √3

According to the question,

Sin A = ½ … (1)

We know that,

 … (2)

To find the value of cos A.

We have the equation,

sin² θ +cos² θ =1

So, cos θ = √(1-sin² θ)

Then,

cos A = √(1-sin² A) … (3)

cos² A = 1-sin² A

cos A = √ (1-sin² A)

Substituting equation 1 in 3, we get,

cos A = √(1-1/4) = √(3/4) = √3/2

Substituting values of sin A and cos A in equation 2, we get

cot A = (√3/2) × 2 = √3

(B) 0

According to the question,

We have to find the value of the equation,

cosec(75°+θ) – sec(15°-θ) – tan(55°+θ) + cot(35°-θ)

= cosec[90°-(15°-θ)] – sec(15°-θ) – tan(55°+θ) + cot[90°-(55°+θ)]

Since, cosec (90°- θ) = sec θ

And, cot(90°-θ) = tan θ

We get,

= sec(15°-θ) – sec(15°-θ) – tan(55°+θ) + tan(55°+θ)

= 0

(C) √(b² – a²)/b

According to the question,

sin θ =a/b

We know, sin² θ +cos² θ =1

sin² A = 1-cos² A

sin A = √(1-cos² A

So, cos θ = √(1-a²/b² ) = √((b²-a²)/b² ) = √(b²-a² )/b

Hence, cos θ = √(b² – a² )/b

(B) cos 2β

According to the question,

cos(α+β) = 0

Since, cos 90° = 0

We can write,

cos(α+β)= cos 90°

By comparing cosine equation on L.H.S and R.H.S,

We get,

(α+β)= 90°

α = 90°-β

Now we need to reduce sin (α -β ),

So, we take,

sin(α-β) = sin(90°-β-β) = sin(90°-2β)

sin(90°-θ) = cos θ

So, sin(90°-2β) = cos 2β

Therefore, sin(α-β) = cos 2β

(B) 1

tan 1°. tan 2°.tan 3° …… tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.tan 45°.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

Since, tan 45° = 1,

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan(90°-44°).tan(90°-43°)…tan(90°-3°). tan(90°-2°).tan(90°-1°)

Since, tan(90°-θ) = cot θ,

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.cot 44°.cot 43°…cot 3°.cot 2°.cot 1°

Since, tan θ = (1/cot θ)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1. (1/tan 44^o). (1/tan 43^o)… (1/tan 3^o). (1/tan 2^o). (1/tan 1^o)

= 1

Hence, tan 1°.tan 2°.tan 3° …… tan 89° = 1

(C) 1

According to the question,

cos 9∝ = sin ∝ and 9∝<90°

i.e. 9α is an acute angle

We know that,

sin(90°-θ) = cos θ

So,

cos 9∝ = sin (90°-∝)

Since, cos 9∝ = sin(90°-9∝) and sin(90°-∝) = sin∝

Thus, sin (90°-9∝) = sin∝

90°-9∝ =∝

10∝ = 90°

∝ = 9°

Substituting ∝ = 9° in tan 5∝, we get,

tan 5∝ = tan (5×9) = tan 45° = 1

∴, tan 5∝ = 1