Question

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

Answer 1

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

$\Rightarrow R.H.S=\frac{\sec \theta +1}{\sec \theta -1}$

$=\frac{(\sec \theta +1)}{(\sec \theta -1)}\times \frac{(\sec \theta +1)}{(\sec \theta +1)}$

$=\frac{{(\sec \theta +1)}^2}{{\sec }^2\theta -1}$

$=\frac{{\sec }^2\theta +1+2\sec \theta }{{\tan }^2\theta }$

$=\frac{{\sec }^2\theta }{{\tan }^2\theta }+\frac{1}{{\tan }^2\theta }+\frac{2\sec \theta }{{\tan }^2\theta }$

$=\frac{1}{{\cos }^2\theta }\times \frac{{\cos }^2\theta }{{\sin }^2\theta }+{\cot }^2\theta +\frac{2}{\cot \theta }\times \frac{{\cos }^2\theta }{{\sin }^2\theta }$

$=cose{c}^2\theta +{\cot }^2\theta +2\cot \theta cosec\theta$

$={(cosec\theta +\cot \theta )}^2$

$=L.H.S$

Hence, the answer is proved.