Question

The value of $\frac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}$ is equal to

• A. $\frac{\cos B+\sin B}{\cos B-\sin B}$
• B. $\frac{\cos A+\sin A}{\cos A-\sin A}$
• C. $\frac{\cos A-\sin A}{\cos A+\sin A}$
• D. $\frac{\sin A-\cos A}{\cos A+\sin A}$

Answer 1

The correct option is B $\frac{\cos A+\sin A}{\cos A-\sin A}$

Let $k=\frac{sin(B+A)+cos(B-A)}{sin(B-A)+cos(B+A)}$

By componendo-dividendo Rule
$\frac{k+1}{k-1}=\frac{sin(A+B)+sin(B-A)+cos(B-A)+cos(B+A)}{sin(A+B)-sin(B-A)+cos(B-A)-cos(B+A)}$
$=\frac{sinBcosA+cosBcosA}{cosBsinA-sinBsin(-A)}$
$=\frac{cosA(sinB+cosB)}{sinA(sinB+cosB)}$
$\Rightarrow \frac{k+1}{k-1}=\frac{cosA}{sinA}$
$\Rightarrow ksinA+sinA=kcosA-cosA$
$\Rightarrow k=\frac{sinA+cosA}{cosA-sinA}$