Question

The value of ${\sin }^{2}A+{\sin }^2(A+B)-2\sin A\cos B\sin (A+B)$ when $B={45}^{\circ }$ is

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. ${\sin }^{2}A+\frac{1}{\sqrt{2}}$
• D. ${\sin }^{2}A+\frac{1}{2}$

Answer 1

The correct option is B $\frac{1}{2}$

${\sin }^{2}A+{\sin }^2(A+B)-2\sin A\cos B\sin (A+B)$

Putting $B={45}^{\circ }$,
${\sin }^2(A+B)=\frac{(\sin A+\cos A{)}^2}{2}=\frac{1}{2}+\sin A\cos A\cdots \left(1\right)$
$2\sin A\cos B\sin (A+B)=\sin A(\sin A+\cos A)={\sin }^{2}A+\sin A\cos A\cdots \left(2\right)$

Now, using equation $\left(1\right)$ and $\left(2\right)$,
${\sin }^{2}A+{\sin }^2(A+B)-2\sin A\cos B\sin (A+B)={\sin }^{2}A+\frac{1}{2}+\sin A\cos A-({\sin }^{2}A+\sin A\cos A)=\frac{1}{2}$