Question

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

Answer 1

From the properties of triangles,

$\Rightarrow 1+{\tan }^{2}x={\sec }^{2}x$

$\Rightarrow 1+{\cot }^{2}x={\text{cosec}}^{2}x$

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

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

$=1+{\tan }^{2}x+1+{\cot }^{2}x={\sec }^{2}x+cose{c}^{2}x=RHS$

$\therefore LHS=RHS$

Hence proved.