Question

If $tan^{-1}(sin^2 \theta - 2 \,sin\, \theta + 3) + cot^{-1}( 5^{sec^2 y} +1 ) = \frac{\pi}{2}$,, then the value of $cos^2\theta - sin\,\theta$ is equal to

• A. $0$
• B. $-1$
• C. $1$
• D. none of these

Answer 1

Answer: (C)

From the given equation $sin^2 \theta - 2\, sin\, \theta + 3 = 5^{sec^2 \,y} + 1$, we get
$(sin \,\theta - 1)^2 + 2 = 5^{sec^2 y} + 1$
L.H.S. $\le 6$, R.H.S. $\ge 6$
Possible solution is $sin \,\theta = - 1$ when L.H.S. = R.H.S.
$\Rightarrow cos^2\, \theta = 0$
$\Rightarrow cos^2\,\theta - sin\,\theta = 1$