If A + B + C = π, prove that

Solution:

$=\frac{\cos A}{\sin B \sin C}+\frac{\cos B}{\sin C \sin A}+\frac{\cos C}{\sin A \sin B}$

Taking L.C.M

$=\frac{\cos A \sin A+\cos B \sin B+\cos C \sin C}{\sin B \sin C \sin A}$

Multiplying and divide the above equation by 2, we get

$=\frac{2 \cos A \sin A+2 \cos B \sin B+2 \cos C \sin C}{2 \sin B \sin C \sin A}$

Since , sin2A = 2sinAcosA

$=\frac{\sin 2 A+\sin 2 B+\sin 2 C}{2 \sin B \sin C \sin A}$

Now,

$=\sin 2 A+\sin 2 B+\sin 2 C$

$=2 \sin A \cos A+2 \sin (B+C) \cos (B-C)$

since A + B + C = π

$\rightarrow \mathrm{B}+\mathrm{A}=180-\mathrm{C}$

$=2 \sin A \cos A+2 \sin (\pi-A) \cos (B-C)$

$=2 \sin A \cos A+2 \sin A \cos (B-C)$

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

$($ but $\cos A=\cos \{180-(B+C)\}=-\cos (B+C)$

And now using $\cos A-\cos B=2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{-A+B}{2}\right)$

$=2 \sin A\{2 \sin B \sin C\}$

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

Putting the above value in the equation, we get

$=\frac{4 \sin A \sin B \sin C}{2 \sin B \sin C \sin A}$

= 2

= R.H.S