Advertisements
Advertisements
Question
If cos 9α = sinα and 9α < 90°, then the value of tan5α is ______.
Options
`1/sqrt(3)`
`sqrt(3)`
1
0
Advertisements
Solution
If cos 9α = sinα and 9α < 90°, then the value of tan5α is 1.
Explanation:
According to the question,
cos 9α = sin α and 9α < 90°
i.e. 9α is an acute angle
We know that,
sin(90° – θ) = cos θ
So, cos 9α = sin(90° – α)
Since, cos 9α = sin(90° – 9α) and sin(90° – α) = sin α
Thus, sin(90° – 9α) = sin α
90° – 9α = α
10α = 90°
α = 9°
Substituting α = 9° in tan 5α, we get,
tan 5α = tan(5 × 9°)
= tan 45°
= 1
∴ tan 5α = 1
APPEARS IN
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(tan theta)/(1-cot theta) + (cot theta)/(1-tan theta) = 1+secthetacosectheta`
[Hint: Write the expression in terms of sinθ and cosθ]
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : m2 – n2 = a2 – b2
`cot^2 theta - 1/(sin^2 theta ) = -1`a
`sec theta (1- sin theta )( sec theta + tan theta )=1`
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
If x = a sin θ and y = bcos θ , write the value of`(b^2 x^2 + a^2 y^2)`
If a cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p2 − q2
Prove the following identity :
`sec^4A - sec^2A = sin^2A/cos^4A`
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
If sinA + cosA = m and secA + cosecA = n , prove that n(m2 - 1) = 2m
If sinA + cosA = `sqrt(2)` , prove that sinAcosA = `1/2`
Prove that :(sinθ+cosecθ)2+(cosθ+ secθ)2 = 7 + tan2 θ+cot2 θ.
Choose the correct alternative:
sec 60° = ?
Prove that `(cos(90 - "A"))/(sin "A") = (sin(90 - "A"))/(cos "A")`
If tan θ = `7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ......[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
