Question

(Sin Q + Cos Q)/ (Sin Q – Cos Q) + (Sin Q – Cos Q) / (Sin Q + Cos Q) = 2 / (2 Sin2Q – 1).

Answer 1

         

$\huge\boxed{ LHS }$

$\boxed{\frac{\sin Q + \cos Q}{\sin Q -\cos Q}+\frac{\sin Q-\cos Q}{\sin Q+\cos Q}} \\ \\ \\ \boxed{\frac{(\sin Q+\cos Q)^2+(\sin Q-\cos Q)^2}{(\sin Q - \cos Q)(\sin Q+\cos Q)}}\ \ \ \ [ \sf{Like} \ \frac{x+y}{x-y}+\frac{x-y}{x+y}=\frac{(x+y)^2+(x-y)^2}{(x+y)(x-y)}] \\ \\ \\ \boxed{\frac{2(\sin^2Q+\cos^2Q)}{\sin^2Q-\cos^2Q}}\ \ \ \ [(x+y)^2+(x-y)^2=2(x^2+y^2)\ \ ; \ \ (x-y)(x+y)=x^2-y^2] \\ \\ \\ \boxed{\frac{2 \times 1}{\sin^2Q-(1-\sin^2Q)}}\ \ \ \ [\sin^2Q+\cos^2Q=1\ \ ; \ \ \sin^2Q=1-\cos^2Q]$

$\boxed{\frac{2}{\sin^2Q-1+\sin^2Q}} \\ \\ \\ \boxed{\frac{2}{2\sin^2Q-1}} \\ \\ \\ \huge\boxed{ RHS }$

$\sf{Hence proved! \\ \\ \\ Plz mark it as the brainliest. \\ \\ Plz ask me if you've any doubts. \\ \\ \\ Thank you. :-)}$