Question

If $ta{n}^{-1}(si{n}^2\theta -2sin\theta +3)+co{t}^{-1}(5{\sin }^2\theta +1)=\frac{\pi }{2}$, then the value of $2co{s}^2\theta -sin\theta$.

• A. $0$
• B. $-1$
• C. $1$
• D. $noneofthese$

Answer 1

Answer: (C)

$1$

We know,

${\tan }^{-1}\left(x\right)+{\cot }^{-1}\left(x\right)=\frac{\pi }{2}$

Given,

${\tan }^{-1}({\sin }^2\theta -2\sin \theta +3)+{\cot }^{-1}(5{\sin }^2\theta +1)=\frac{\pi }{2}$

$\Rightarrow {\sin }^2\theta -2\sin \theta +3=5{\sin }^2\theta +1$

$\Rightarrow -4{\sin }^2\theta -2\sin \theta +2=0$

$\Rightarrow -2{\sin }^2\theta -\sin \theta +1=0$

$\Rightarrow -2(1-{\cos }^2\theta )-\sin \theta +1=0$

$\Rightarrow -2+2{\cos }^2\theta -\sin \theta +1=0$

$\Rightarrow 2{\cos }^2\theta -\sin \theta -1=0$

$\Rightarrow 2{\cos }^2\theta -\sin \theta =1$