Question

Prove that: $\sqrt{{\sec }^{2}A+\cos e{c}^{2}A}=\tan A+\cot A$

Answer 1

$\sqrt{se{c}^{2}A+cose{c}^{2}A}$

$=\sqrt{1+ta{n}^{2}A+1+co{t}^{2}A}$

$=\sqrt{ta{n}^{2}A+co{t}^{2}A+2}$

$=\sqrt{ta{n}^{2}A+co{t}^{2}A+2tanA.cotA}$

$=\sqrt{(tanA+cotA{)}^2}$

$=tanA+cotA$ (proved)