Question

solve this step by step.
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Answer 1

Answer:

LHS: $\frac{cos(90-x)}{1+sin(90-x))}$  $+$ $\frac{1+sin(90-x)}{cos(90-x)}$

We know that Cos(90-x)= Sin x and Sin(90-x)= Cos x

Therefore, $\frac{sin x}{1+cos x}$ + $\frac{1+cosx}{sinx}$

By taking LCM and multiplying, we get:

           $\frac{sin^2x+ (1+cosx)^2}{(1+cosx)sinx }$=$\frac{sin^2x+cos^2x+1+2cosx}{(1+cosx)sinx}$= $\frac{2+2cosx}{(1+cosx)sinx}$ = $\frac{2(1+cosx)}{(1+cosx)sinx}$

Divide 1+cosx and we get: $\frac{2}{sinx}$ = $2cosecx$

           = RHS

Hence Proved