Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $\cos A+\sin A-\frac{2}{\cos B+\sin B}=0,$ then the value of ${\left(\frac{a+b}{c}\right)}^4$ is?

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

Answer: (A)

$4$

$\cos A+\sin A-\frac{2}{\cos B+\sin B}=0$
$\Rightarrow (\cos A+\sin A)(\cos B+\sin B)=2$
$\Rightarrow \cos (A-B)+\sin (A+B)=2$
$\Rightarrow \cos (A-B)+\sin \left(C\right)=2$
$\therefore A=B=\frac{\pi }{4}$ and $C=\frac{\pi }{2}$
Now, ${\left(\frac{a+b}{c}\right)}^4={\left(\frac{\sin A+\sin B}{\sin C}\right)}^4=4$
Hence, option A.