If the solution of the equation \(\log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1, x \in\left(0, \frac{\pi}{2}\right)\), is \(\sin ^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right)\), where \(\alpha, \beta\) are integers, then \(\alpha+\beta\) is equal to:

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Trigonometric Equations

If the solution of the equation

lo g_{c o s x} cot x + 4 lo g_{s i n x} tan x = 1, x \in (0, \frac{π}{2})

, is

sin^{- 1} (\frac{α + β}{2})

, where

α, β

are integers, then

α + β

is equal to:

Answer: (D)

Writing

u = ln sin x

and

v = ln cos x

, the logarithms convert to

lo g_{c o s x} cot x = \frac{v - u}{v}

and

lo g_{s i n x} tan x = \frac{u - v}{u}

. The equation becomes

\frac{v - u}{v} + \frac{4 ( u - v )}{u} = 1

, which simplifies to

\frac{u}{v} + \frac{4 v}{u} = 4

, i.e.

u^{2} - 4 uv + 4 v^{2} = 0

, so

(u - 2 v)^{2} = 0

and

u = 2 v

. Hence

ln sin x = 2 ln cos x

, giving

sin x = cos^{2} x = 1 - sin^{2} x

, so

sin^{2} x + sin x - 1 = 0

and

sin x = \frac{- 1 + 5}{2}

. Thus

x = sin^{- 1} \frac{- 1 + 5}{2}

, giving

α = - 1, β = 5

and

α + β = 4.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Single correct option