Advertisements
Advertisements
प्रश्न
If cos 9α = sinα and 9α < 90°, then the value of tan5α is ______.
पर्याय
`1/sqrt(3)`
`sqrt(3)`
1
0
Advertisements
उत्तर
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
संबंधित प्रश्न
(secA + tanA) (1 − sinA) = ______.
Prove the following identities:
`cotA/(1 - tanA) + tanA/(1 - cotA) = 1 + tanA + cotA`
`(1 + cot^2 theta ) sin^2 theta =1`
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
`(cos ec^theta + cot theta )/( cos ec theta - cot theta ) = (cosec theta + cot theta )^2 = 1+2 cot^2 theta + 2cosec theta cot theta`
If m = ` ( cos theta - sin theta ) and n = ( cos theta + sin theta ) "then show that" sqrt(m/n) + sqrt(n/m) = 2/sqrt(1-tan^2 theta)`.
Write the value of `( 1- sin ^2 theta ) sec^2 theta.`
Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
\[\frac{x^2 - 1}{2x}\] is equal to
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
Prove that:
`sqrt((sectheta - 1)/(sec theta + 1)) + sqrt((sectheta + 1)/(sectheta - 1)) = 2cosectheta`
Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
If 1 – cos2θ = `1/4`, then θ = ?
Prove that `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ
Prove that `sec"A"/(tan "A" + cot "A")` = sin A
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
(sec2 θ – 1) (cosec2 θ – 1) is equal to ______.
