Question

Prove tan*2A +cot*2A+2=sec*A.cosec*2A

Answer 1

Here $LHS=\tan ^2 A+\cot^2 A$. Now use the formula,

$x^4+y^4=(x^2+y^2)^2-2x^2y^2$

$\sin^2 A+\cos ^2 A=1$

$LHS=\tan ^2 A+\cot^2 A+2\\ LHS=\frac{\sin^2 A}{\cos^2 A}+\frac{\cos^2 A}{\sin^2 A}+2\\ LHS=\frac{\sin^4 A +\cos^4 A}{\sin^2 A\cos^2 A}+2\\ LHS=\frac{(\sin^2 A +\cos^2 A)^2-2\sin^2 A \cos^2 A}{\sin^2 A\cos^2 A}+2\\ LHS=\frac{(1)^2-2\sin^2 A \cos^2 A}{\sin^2 A\cos^2 A}+2$

$LHS=\frac{1}{\sin^2 A\cos^2 A}-2+2\\ LHS=\sec^2 A\csc^2 A=RHS$

Proof is complete.