Question

Prove that
$\frac{sin({90}^o-\theta )}{cosec({90}^o-\theta )-cot({90}^o-\theta )}=1+sin\theta$

Answer 1

$\sin (90°-\theta )=\cos \theta$

$\cot (90°-\theta )=\tan \theta$

$\csc (90°-\theta )=\sec \theta$

Thus the question becomes: $\frac{\cos \theta }{\sec \theta -\tan \theta }$

$=\frac{\cos \theta }{\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }}$

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

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

$=(1+\sin \theta )$