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KCET · Maths · Trigonometric Ratios & Identities

\(\sin ^{2} 17.5^{\circ}+\sin ^{2} 72.5^{\circ}\) is equal to

  1. A \(\cos ^{2} 90^{\circ}\)
  2. B \(\tan ^{2} 45^{\circ}\)
  3. C \(\cos ^{2} 30^{\circ}\)
  4. D \(\sin ^{2} 45^{\circ}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\tan ^{2} 45^{\circ}\)

Step-by-step Solution

Detailed explanation

We have, \(\sin ^{2} 17.5^{\circ}+\sin ^{2} 72.5^{\circ}\)
\[
\begin{aligned}
&=\sin ^{2} 17.5^{\circ}+\sin ^{2}\left(90^{\circ}-17.5^{\circ}\right) \\
&=\sin ^{2} 17.5^{\circ}+\cos ^{2} 17.5^{\circ} \\
&=1=1^{2}=\tan ^{2} 45^{\circ}
\end{aligned}
\]