Question

If A,B,C are angles of a triangle, then $2\sin \frac{A}{2}cosec\frac{B}{2}\sin \frac{C}{2}-\sin A\cot \frac{B}{2}-\cos A$ is

• A. independent of $A,B,C$
• B. function of $A,B$
• C. function of $C$
• D. none of these

Answer 1

Answer: (A)

independent of $A,B,C$

$A+B+C=\pi$
$k=2\sin \frac{A}{2}\csc \frac{B}{2}\sin \frac{C}{2}-\sin A\cot \frac{B}{2}-\cos A$
$\Rightarrow k=\frac{2\sin \frac{A}{2}\sin \frac{C}{2}-2\sin \frac{A}{2}\cos \frac{A}{2}\cos \frac{B}{2}}{\sin \frac{B}{2}}-\cos A$
$\Rightarrow k=\frac{2\sin \frac{A}{2}[\sin \left(\frac{\pi -A-B}{2}\right)-\frac{1}{2}(\cos \left(\frac{A+B}{2}\right)+\cos \left(\frac{A-B}{2}\right))]}{\sin \frac{B}{2}}-\cos A$
$\Rightarrow k=\frac{2\sin \frac{A}{2}[\cos \left(\frac{A+B}{2}\right)-\frac{1}{2}(\cos \left(\frac{A+B}{2}\right)+\cos \left(\frac{A-B}{2}\right))]}{\sin \frac{B}{2}}-\cos A$
$\Rightarrow k=\frac{\sin \frac{A}{2}[\cos \left(\frac{A+B}{2}\right)-\cos \left(\frac{A-B}{2}\right)]}{\sin \frac{B}{2}}-\cos A$
$\Rightarrow k=\frac{-2\sin \frac{A}{2}\left[\sin \frac{1}{2}\left(\frac{A+B+A-B}{2}\right)\sin \frac{1}{2}\left(\frac{A+B-A+B}{2}\right)\right]}{\sin \frac{B}{2}}-\cos A$
$\Rightarrow k=\frac{-2\sin \frac{A}{2}\sin \frac{A}{2}\sin \frac{B}{2}}{\sin \frac{B}{2}}-(1-2{\sin }^2\frac{A}{2})$
$\Rightarrow k=-2{\sin }^2\frac{A}{2}-(1-2{\sin }^2\frac{A}{2})=-1$
Therefore k is independent of A,B,C
Ans: A