Question

Is LHS=RHS?
${\left\{\frac{1+\cos \theta +\sin \theta }{1+\cos \theta -\sin \theta }\right\}}^2=\frac{1+\sin \theta }{1-\sin \theta },1-\cos \theta \ne 0$

• A. Yes
• B. No
• C. Ambiguos
• D. Data insufficient

Answer 1

The correct option is A Yes

Given

LHS

${\left\{\frac{1+\cos \theta +\sin \theta }{1+\cos \theta -\sin \theta }\right\}}^2$

$\frac{1+{\cos }^2\theta +{\sin }^2\theta +2\cos \theta +2\sin \theta +2\cos \theta \sin \theta }{1+{\cos }^2\theta +{\sin }^2\theta +2\cos \theta -2\sin \theta -2\cos \theta \sin \theta }$

$\frac{2+2\cos \theta +2\sin \theta +2\cos \theta \sin \theta }{2+2\cos \theta -2\sin \theta -2\cos \theta \sin \theta }$

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

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

$⟹\frac{1+\sin \theta }{1-\sin \theta }$