If cos 9α = sinα and 9α < 90° , then the value of tan5α is(A) 1/�

1.	If cos 9α = sinα and 9α < 90° , then the value of tan5α is(A) 1/√3 (B) √3 (C) 1 (D) 0
Answer» (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

If cos 9α = sinα and 9α < 90° , then the value of tan5α is(A) 1/√3 (B) √3 (C) 1 (D) 0

Answer»

(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