MHT CET · Maths · Trigonometric Ratios & Identities
If \(A\) and \(B\) are supplementary angles, then \(\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}=\)
- A 1
- B \(\frac{1}{3}\)
- C 0
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
(C)
\(A\) and \(B\) are supplementary angles. \(\Rightarrow A+B=180 \Rightarrow A=180-B \Rightarrow \frac{A}{2}=90-\frac{B}{2}\)
\(\therefore \sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}=\sin ^{2} \frac{A}{2}+\sin ^{2}\left(90-\frac{A}{2}\right)=\sin ^{2} \frac{A}{2}+\) \(\cos ^{2}\left(\frac{A}{2}\right)=1\)
\(A\) and \(B\) are supplementary angles. \(\Rightarrow A+B=180 \Rightarrow A=180-B \Rightarrow \frac{A}{2}=90-\frac{B}{2}\)
\(\therefore \sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}=\sin ^{2} \frac{A}{2}+\sin ^{2}\left(90-\frac{A}{2}\right)=\sin ^{2} \frac{A}{2}+\) \(\cos ^{2}\left(\frac{A}{2}\right)=1\)
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