Question

Find the value of $x$ from the given equation $\csc ({90}^o-\theta )+x\cos \theta .\cot ({90}^o-\theta )=\sin ({90}^o-\theta )$

• A. $\tan \theta$
• B. $-\tan \theta$
• C. $-\cot \theta$
• D. None of these

Answer 1

The correct option is B $-\tan \theta$

$cosec({90}^0-\theta )+x\cos \theta \cot ({90}^0-\theta )=\sin ({90}^0-\theta )$

$\Rightarrow \frac{1}{\sin ({90}^0-\theta )}+x\cos \theta \frac{\cos ({90}^0-\theta )}{\sin ({90}^0-\theta )}=\sin ({90}^0-\theta )$

$\Rightarrow 1+x\cos \theta \cos ({90}^0-\theta )={\sin }^2({90}^0-\theta )$

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

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

$\Rightarrow x=-\frac{\sin \theta }{\cos \theta }=-\tan \theta$ Ans.