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Question
Solve: `2tan^-1 (cos x) = tan^-1 (2"cosec" x)`
Solution
`2tan^-1 (cos x) = tan^-1 (2"cosec" x)`
`2tan^-1x = tan^-1 [(2x)/(1 - x^2)]`
⇒ `2tan^-1 (cos x) = tan^-1 [(2cosx)/(1 - cos^2x)]`
`1tan^-1 (cosx) = tan^-1[(2cosx)/(sin^2x)]`
Now, `tan^-1 (2 "cosec" x) = tan^-1 [(2cosx)/(sin^2x)]`
⇒ `2 "cosec" x = (2cosx)/(sin^2x)`
`1/sinx = cosx/sinx`
sin2x = sin x cos x
⇒ sin x cos x – sin2x = 0
⇒ sin x(cos x – sin x) = 0
sin x = 0 or cos x – sin x = 0
⇒ x = nπ, n ∈ z, or cos x = sin x
tan x = 1 = `tan pi/4`
⇒ x = `"n"pi + pi/4`, n ∈ z
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