Question

In a $△ABC$, if $(\sin A+\cos A)(\sin B+\cos B)=2$, then the value of $1+{\sec }^2\left(\frac{C}{2}\right)$ is

Answer 1

$(\sin A+\cos A)(\sin B+\cos B)=2\Rightarrow \sin A\sin B+\sin A\cos B+\cos A\sin B+\cos A\cos B=2\Rightarrow \cos A\cos B+\sin A\sin B+\cos A\sin B+\sin A\cos B=2\Rightarrow \cos (A-B)+\sin (A+B)=2$
We know that the maximum value of sine and cosine functions is $1$, so
$\cos (A-B)=1\text{and}\sin (A+B)=1\Rightarrow A-B=0\text{and}A+B=\frac{\pi }{2}\Rightarrow A=B=\frac{\pi }{4}\Rightarrow C=\frac{\pi }{2}\Rightarrow 1+{\sec }^2\left(\frac{C}{2}\right)=1+{\sec }^2\left(\frac{\pi }{4}\right)=3$