Question

Cos(90-theta)/1+sin(90-theta)+1+sin(90-theta)/cos(90-theta)=2cosectheta

Answer 1

Answer:

Step-by-step explanation:

Ok it's done

Answer 2

∴ Hence proved.

Step-by-step explanation:

Given;

Proved that  $\frac{cos(90-\theta)}{1+sin(90-\theta)} +\frac{1+sin(90-\theta)}{cos(90-\theta)} =2cosec\theta$

∴ LHS:

$\frac{cos(90-\theta)}{1+sin(90-\theta)} +\frac{1+sin(90-\theta)}{cos(90-\theta)}$

$=\frac{sin\theta}{1+cos\theta} +\frac{1+cos\theta}{sin\theta}$   (∵$cos(90-\theta)=sin\theta$ and $sin(90-\theta)=cos\theta$ )

$=\frac{sin^{2} \theta+(1+cos\theta)^{2} }{sin\theta(1+cos\theta)}$

$=\frac{sin^{2} \theta+1+2cos\theta +cos^{2} \theta }{sin\theta(1+cos\theta)}$

$=\frac{1+1+2cos\theta}{sin\theta(1+cos\theta)}$  (∵ $sin^{2} \theta+cos^{2} \theta=1$ )

$=\frac{2(1+cos\theta)}{sin\theta(1+cos\theta)}$

$=\frac{2}{sin\theta}$   (By cancel $(1+cos\theta)$ in numerator and denominator )

$=2cosec \theta$ (∵ $\frac{1}{sin\theta} =cosec\theta$ )

$=$RHS

∴ Hence proved.