Question

Prove that $\frac{\sin ({90}^o-\theta )}{1+\sin \theta }+\frac{\cos \theta }{1-\cos ({90}^o-\theta )}=2\sec \theta$

Answer 1

LHS $=\frac{\sin ({90}^o-\theta )}{1+\sin \theta }+\frac{\cos \theta }{1-\cos ({90}^o-\theta )}$
$=\frac{\cos \theta }{1+\sin \theta }+\frac{\cos \theta }{1-\sin \theta }$
$=\frac{\cos \theta (1-\sin \theta )+\cos \theta (1+\sin \theta )}{1-{\sin }^2\theta }=\frac{\cos \theta -\cos \theta \sin \theta +\cos \theta +\cos \theta \sin \theta }{1-{\sin }^2\theta }$
$=\frac{2\cos \theta }{{\cos }^2\theta }=\frac{2}{\cos \theta }=2\sec \theta$
$=2\sec \theta =$ RHS