Question

if $\frac{2\sin A}{1+\sin A+\cos A}=k,$ then $\frac{1+\sin A-\cos A}{1+\sin A}=$

• A. $\frac{k}{2}$
• B. $k$
• C. $2k$
• D. $\frac{1}{k}$

Answer 1

The correct option is B $k$

$(1+sinA+cosA)(1+sinA-cosA)={(1+sinA)}^2-{\cos }^{2}A=2sinA(1+sinA)$

$=\frac{2sinA}{1+sinA+cosA}=\frac{1+sinA-cosA}{1+sinA}=\frac{1+sinA-cosA}{1+sinA}=k$ [B]