The general solution of \tan x- \sin x=1-\tan x \cdot \sin x

Correct Answer is : x=nπ+π4 and or x=nπ+(−1)a(−π2)

x=n \pi+\frac{\pi}{4} and or x=n \pi+(-1)^a (\frac{-\pi}{2})

x=\frac{n \pi}{4}-\frac{\pi}{4} and or x=n \pi+(-1)^n (-\frac{\pi}{2})

x=n \pi+\frac{\pi}{6} and or \pi=n \pi+(-1)^n (-\frac{\pi}{2})

COMEDK Exam 2012 Solved Paper

Question : 177 of 180

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The general solution of tan‌x−sin‌x=1−tan‌x⋅sin‌x

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