Question

Find $a$ if $\frac{\cos \alpha }{{\cot }^2\frac{\alpha }{2}-{\tan }^2\frac{\alpha }{2}}=\frac{1}{a}{\sin }^2\alpha .$

Answer 1

$\frac{\cos \alpha }{{\cot }^2\frac{\alpha }{2}-{\tan }^2\frac{\alpha }{2}}=\frac{\cos \alpha }{\frac{{\cos }^4\frac{\alpha }{2}-{\sin }^4\frac{\alpha }{2}}{{\cos }^2\frac{\alpha }{2}{\sin }^2\frac{\alpha }{2}}}=\frac{\cos \alpha .{\cos }^2\frac{\alpha }{2}{\sin }^2\frac{\alpha }{2}}{{\cos }^4\frac{\alpha }{2}-{\sin }^4\frac{\alpha }{2}}=\frac{1}{4}\frac{\cos \alpha .4{\cos }^2\frac{\alpha }{2}{\sin }^2\frac{\alpha }{2}}{{\cos }^4\frac{\alpha }{2}-{\sin }^4\frac{\alpha }{2}}=\frac{1}{4}\frac{\cos \alpha .{\sin }^2\alpha }{\cos \alpha }=\frac{1}{4}{\sin }^2\alpha .$

Hence , Value of $a=4$

Formula Used:

$4{\cos }^2\frac{\alpha }{2}{\sin }^2\frac{\alpha }{2}={\sin }^2\alpha {\cos }^4\frac{\alpha }{2}-{\sin }^4\frac{\alpha }{2}=({\cos }^2\frac{\alpha }{2}+{\sin }^2\frac{\alpha }{2})({\cos }^2\frac{\alpha }{2}-{\sin }^2\frac{\alpha }{2})=1.({\cos }^2\frac{\alpha }{2}-{\sin }^2\frac{\alpha }{2})=\cos \alpha$