KCET · Maths · Trigonometric Ratios & Identities
\(\sin ^{2} 17.5^{\circ}+\sin ^{2} 72.5^{\circ}\) is equal to
- A \(\cos ^{2} 90^{\circ}\)
- B \(\tan ^{2} 45^{\circ}\)
- C \(\cos ^{2} 30^{\circ}\)
- D \(\sin ^{2} 45^{\circ}\)
Answer & Solution
Correct Answer
(B) \(\tan ^{2} 45^{\circ}\)
Step-by-step Solution
Detailed explanation
We have, \(\sin ^{2} 17.5^{\circ}+\sin ^{2} 72.5^{\circ}\)
\[
\begin{aligned}
&=\sin ^{2} 17.5^{\circ}+\sin ^{2}\left(90^{\circ}-17.5^{\circ}\right) \\
&=\sin ^{2} 17.5^{\circ}+\cos ^{2} 17.5^{\circ} \\
&=1=1^{2}=\tan ^{2} 45^{\circ}
\end{aligned}
\]
\[
\begin{aligned}
&=\sin ^{2} 17.5^{\circ}+\sin ^{2}\left(90^{\circ}-17.5^{\circ}\right) \\
&=\sin ^{2} 17.5^{\circ}+\cos ^{2} 17.5^{\circ} \\
&=1=1^{2}=\tan ^{2} 45^{\circ}
\end{aligned}
\]
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