Question

If $\frac{2\sin \alpha }{1+\sin \alpha +\cos \alpha }=\lambda$, then what will be the value of $\frac{1+\sin \alpha -\cos \alpha }{1+\sin \alpha }$

• A. $2\lambda$
• B. $\lambda$
• C. $-\lambda$
• D. $0$

Answer 1

Answer: (B)

$\lambda$

Consider the given equation.

$\frac{2\sin \alpha }{1+\sin \alpha +\cos \alpha }=\lambda$ ……. (1)

$\frac{2\sin \alpha }{1+\sin \alpha +\cos \alpha }\times \frac{(1+\sin \alpha -\cos \alpha )}{1+\sin \alpha -\cos \alpha }=\lambda$

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

$\frac{2\sin \alpha +2{\sin }^2\alpha -2\sin \alpha \cos \alpha }{1+{\sin }^2\alpha +2\sin \alpha -{\cos }^2\alpha }=\lambda$

$\frac{2\sin \alpha +2{\sin }^2\alpha -2\sin \alpha \cos \alpha }{{\sin }^2\alpha +2\sin \alpha +1-{\cos }^2\alpha }=\lambda$

$\frac{2\sin \alpha +2{\sin }^2\alpha -2\sin \alpha \cos \alpha }{{\sin }^2\alpha +2\sin \alpha +{\sin }^2\alpha }=\lambda [\because {\sin }^{2}x=1-{\cos }^{2}x]$

$\frac{2\sin \alpha +2{\sin }^2\alpha -2\sin \alpha \cos \alpha }{2{\sin }^2\alpha +2\sin \alpha }=\lambda$

$\frac{1+\sin \alpha -\cos \alpha }{1+\sin \alpha }=\lambda$

Hence, this is the answer.