Question

$\mathrm{In}△ABC\mathrm{if}\cos A+\sin A-\frac{2}{\cos B+\sin B}=0,\mathrm{th}en\mathrm{the}\mathrm{triangle}\mathrm{is}\mathrm{an}$

• A. Equilateral triangle
• B. Isoceles triangle
• C. Scalene triangle

Answer 1

The correct option is B Isoceles triangle

For the given triangle, we have
$\cos A+\sin B=\frac{2}{\cos B+\sin B}\Rightarrow \left(\cos A+\sin B\right)\left(\cos B+\sin B\right)=2\Rightarrow \cos A\cos B+\cos A\sin B+\sin B\cos B+\sin B\sin B=2$
Now, we know the expressions:$\cos (A-B)=\cos A\cos B+\sin A\sin B\sin (A+B)=\sin A\cos B+\sin B\cos A$
Now, using these two expressions, we get
$\cos A\cos B+\cos A\sin B+\sin B\cos B+\sin B\sin B=2\Rightarrow \cos (A-B)+\sin (A+B)=2$
Now, the above expression is possible only if $\cos (A-B)=1\mathrm{and}\sin (A+B)=1$
Since, both sine and cosine ranges from -1 to 1.
Thus, $\cos (A-B)=1\Rightarrow A-B=0\Rightarrow A=B$
Also, $\sin (A+B)=1\Rightarrow A+B=\frac{\pi }{2}$
Now, substituting $A=B$ in the above equation, we get
$A=B=\frac{\pi }{4}.$
Thus, the triangle ABC is an isoceles triangle.