Question

prove the identity:
sinA/1+cosA+1+cosA/sinA= 2cosecA​

Answer 1

Answer:

$\frac{ SinA}{(1 + CosA)} +\frac{ (1 + CosA)}{sinA}$

= $\frac{(SinA)^2 +(1+CosA)^2}{SinA(1 + CosA)}$

= $\frac{(SinA)^2 + (CosA)^2 + 1 + 2CosA}{[SinA(1 + CosA)}$

=$\frac{[2(1 + CosA)]}{[SinA(1 + CosA)]}$

${since, (SinA)^2 + (CosA)^2 = 1 }$

=$\frac{2}{SinA }$

= $2cosesA$

Hence L.H.S=R.H.S.