Question

If $\frac{2\sin \alpha }{1+\sin \alpha +\cos \alpha }=\lambda$, then $\frac{1+\sin \alpha -\cos \alpha }{1+\sin \alpha }$ is equal to

• A. $\frac{1}{\lambda }$
• B. $\lambda$
• C. $1-\lambda$
• D. $1+\lambda$

Answer 1

Answer: (B)

$\lambda$

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

Multiplying numerator and denominator by $(1+\sin \alpha +\cos \alpha )$, we get

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

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

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