Question

If ${\left(\frac{1+\sin \theta -\cos \theta }{1+\sin \theta +\cos \theta }\right)}^2=\lambda \left(\frac{1-\cos \theta }{1+\cos \theta }\right)$, then $\lambda$ equals

• A. $-1$
• B. $1$
• C. $2$
• D. $-2$

Answer 1

The correct option is B $1$

${\left(\frac{1+\sin \theta -\cos \theta }{1+\sin \theta +\cos \theta }\right)}^2=\lambda \left(\frac{1-\cos \theta }{1+\cos \theta }\right)$

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

$=\left[\frac{1+1+2(\sin \theta -\cos \theta )-\sin 2\theta }{1+1+2(\sin \theta -\cos \theta )+\sin 2\theta }\right]=$

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

$=\frac{(1+\sin \theta )}{1+\sin \theta )}\left[\frac{1-\cos \theta }{1+\cos \theta }\right]\Rightarrow \lambda =1$