Question

If $\pi <\alpha <2\pi$ then $\frac{1}{\sin \alpha -\sqrt{{\cot }^2\alpha -{\cos }^2\alpha }}=$

• A. $\sin \alpha$
• B. $-\sin \alpha$
• C. $\frac{1}{\sin \alpha }$
• D. $1$

Answer 1

The correct option is A $\sin \alpha$

According to question,

$\frac{1}{sin\alpha -\sqrt{\frac{co{s}^2\alpha }{si{n}^2\alpha }-co{s}^2\alpha }}$

$⟹\frac{1}{sin\alpha -\sqrt{co{s}^2\alpha (\frac{1}{si{n}^2\alpha }-1)}}$

$⟹\frac{1}{sin\alpha -\sqrt{\frac{co{s}^4\alpha }{si{n}^2\alpha }}}$

$⟹\frac{1}{sin\alpha -|\frac{co{s}^2\alpha }{sin\alpha }|}$

$⟹\frac{1}{sin\alpha +\frac{co{s}^2\alpha }{sin\alpha }}$ ,.......As $(\pi <\alpha <2\pi )$

$⟹\frac{sin\alpha }{si{n}^2\alpha +co{s}^2\alpha }$

$⟹sin\alpha$