Question

$A,B,C$ are interior angles of $ABC$. Prove that $\text{cosec}\left(\frac{A+B}{2}\right)=\sec \frac{C}{2}$ .

Answer 1

$LHS=cosec\left(\frac{A+B}{2}\right)$

$=\frac{1}{sin\left(\frac{A+B}{2}\right)}$ In $\Delta ABC$,

$=\frac{1}{sin(90-\frac{C}{2})}$ $A+B+C=180$

$\frac{1}{sin(90-\frac{c}{2})}$ $A+B=180-C$

$=\frac{1}{cos\left(\frac{C}{2}\right)}$ $\frac{A+B}{2}=\frac{180-C}{2}$

$=sec\left(\frac{C}{2}\right)$ $\frac{A+B}{2}=90-\frac{C}{2}$ Taking sines

$=sec\left(\frac{C}{2}\right)$ $sin\left(\frac{A+B}{2}\right)=sin(90-\frac{C}{2})$

$=RHS$ $=cos\left(\frac{C}{2}\right)$

$\because sin(90-\theta )=cos\theta$

$cos\theta =\frac{1}{sec\theta }$