Question

${\left(\tan x+\cot x\right)}^2={\sec }^{2}x+\cos e{c}^{2}x$

Answer 1

$LHS=(\tan x+\cot x{)}^2$

$={\tan }^{2}x+{\cot }^{2}x+2\tan x\cot x$

$=\frac{{\sin }^{2}x}{{\cos }^{2}x}+\frac{{\cos }^{2}x}{{\sin }^{2}x}+2$ [since we know $\tan x=\frac{1}{cotx},\tan x=\frac{\sin x}{\cos x},\cot x=\frac{\cos x}{\sin x}$]

$=\frac{{\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}$

$=\frac{({\sin }^{2}x+{\cos }^{2}x{)}^2}{{\sin }^{2}x{\cos }^{2}x}$ [\sin ce ${\sin }^{2}x+{\cos }^{2}x=1$]

$=\frac{1}{{\sin }^{2}x}\frac{1}{{\cos }^{2}x}$

$={\sec }^{2}x\cos e{c}^{2}x$

Hence,

$LHS=RHS.$