Question

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

Answer 1

$LHS=(\sin \theta +\csc \theta {)}^2+(\cos \theta +\sec \theta {)}^2$

$={\sin }^{2}A+{\csc }^{2}A+2\sin A\csc A+{\cos }^{2}A+{\sec }^{2}A+2\cos A\sec A$

$=({\sin }^{2}A+{\cos }^{2}A)+2\sin A\times \frac{1}{\sin A}+2\cos A\times \frac{1}{\cos A}+{\csc }^{2}A+{\sec }^{2}A$

$=({\sin }^{2}A+{\cos }^{2}A)+2+2+{\csc }^{2}A+{\sec }^{2}A$

$=1+2+2+(1+{\cot }^{2}A)+(1+{\tan }^{2}A)$ $[\because {\sec }^{2}A=1+{\tan }^A,{\csc }^{2}A=1+{\cot }^{2}A]$

$=7+{\cot }^{2}A+{\tan }^{2}A$

$=RHS$