Question

${(\sin \theta +cosec\theta )}^2+{(\cos \theta +\sec \theta )}^2$ = $7$ +${\tan }^2\theta +{\cot }^2\theta$

Answer 1

Given, ${(\sin \theta +cosec\theta )}^2+{(\cos \theta +\sec \theta )}^2=$L.H.S

$={\sin }^2\theta +cose{c}^2\theta +2\sin \theta .cosec\theta +{\cos }^2\theta +{\sec }^2\theta +2\cos \theta .\sec \theta$

We know that,

$cosec\theta =\frac{1}{\sin \theta }$ & $\sec \theta =\frac{1}{\cos \theta }$

$\Rightarrow {\sin }^2\theta +cose{c}^2\theta +2+{\cos }^2\theta +{\sec }^2\theta +2$

$\Rightarrow {\sin }^2\theta +{\cos }^2\theta +cose{c}^2\theta +{\sec }^2\theta +4$

$\Rightarrow 1+4+{\sec }^2\theta +cose{c}^2\theta$

$\Rightarrow 5+(1+{\tan }^2\theta )+(1+{\cot }^2\theta )$

$\Rightarrow 7+{\tan }^2\theta +{\cot }^2\theta =$R.H.S

Hence proved.