Question

Prove that$\frac{1}{2+{\cot }^2(-\theta )}=\frac{1}{2{\text{cosec}}^2(-\theta )-{\cot }^2(-\theta )}$

Answer 1

Consider LHS$=\frac{1}{2+\cot {}^2(-\theta )}$
Since, $\text{cosec}{}^2(-\theta )-\cot {}^2(-\theta )=1$
Substitute above formula in denominator of given expression, we get
$=\frac{1}{2[\text{cosec}{}^2(-\theta )-\cot {}^2(-\theta )]+\cot {}^2(-\theta ))}$
$=\frac{1}{2\text{cosec}{}^2(-\theta )-2\cot {}^2(-\theta )+\cot {}^2(-\theta ))}$
$=\frac{1}{2\text{cosec}{}^2(-\theta )-\cot {}^2(-\theta )}$
$=$RHS
Hence, proved .