If $A$ and $B$ are supplementary angles, then $\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}=$

If $A$ and $B$ are supplementary angles, then $\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}=$
  1. 1
  2. $\frac{1}{3}$
  3. 0
  4. $\frac{1}{2}$

Solution

$A$ and $B$ are supplementary angles. $\Rightarrow A+B=180 \Rightarrow A=180-B \Rightarrow \frac{A}{2}=90-\frac{B}{2}$ $\therefore \sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}=\sin ^{2} \frac{A}{2}+\sin ^{2}\left(90-\frac{A}{2}\right)=\sin ^{2} \frac{A}{2}+\cos ^{2}\left(\frac{A}{2}\right)=1$

Asked in: MHT CET 2020 (19 Oct Shift 2)

Practice more Trigonometric Ratios & Identities questions on Aicharya