Question

If in a traingle $ABC$, $4\cos A\cos B+\sin 2A+\sin 2B+\sin 2C=4$, then the trangle is

• A. Equilateral
• B. Only right angle
• C. Isosceles & right angle
• D. Isoscleles Triangle

Answer 1

The correct option is C Isosceles & right angle

$△ABC,\sin 2A+\sin 2B+\sin 2C$$=4\sin A\sin B\sin C$

and
$=1+{\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}c=-4\cos A\cos B\cos C$

$=4\cos A\cos B+4\sin A\sin B\sin C=4$

$=\cos A\cos B+\sin A\sin B\sin C=1$

checking for equilateral triangle $\therefore A=B=C=\frac{A}{3}$

$\Rightarrow \frac{1}{4}+\frac{\frac{3}{3}}{8}\ne 1\therefore \Delta$ is not equilateral triangle for right angle put $c={90}^{\circ }$

$\therefore \cos A\cos B+\sin A\sin B=1$

$\therefore \cos (A-B)=1,A=B\Rightarrow$isosceles right angles.

$\therefore A=B\therefore {\cos }^{2}A+{\sin }^{2}A\sin C=1$

The only possible solution is sin $C=1$

i.e $C={90}^{\circ }$

$\therefore$ the triangles should be isosceles right-angle triangles.
Hence, option 'C' is correct.