AP EAMCET · Maths · Trigonometric Ratios & Identities
\(\begin{aligned}
& \sin ^2 18^{\circ}+\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\sin ^2 42^{\circ}+\sin ^2 78^{\circ}+ \\
& \sin ^2 90^{\circ}+\sin ^2 96^{\circ}+\sin ^2 102^{\circ}+\sin ^2 138^{\circ}+\sin ^2 162^{\circ}=
\end{aligned}\)
- A \(\frac{11}{2}\)
- B \(\frac{9}{2}\)
- C 5
- D 4
Answer & Solution
Correct Answer
(A) \(\frac{11}{2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \begin{array}{r}\text { Since, } \sin ^2 18^{\circ}+\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\sin ^2 42^{\circ}+\sin ^2 78^{\circ} \\ +\sin ^2 90^{\circ}+\sin ^2 96^{\circ}+\sin ^2(180-78)^{\circ}+\sin ^2(180-42)^{\circ}+ \\ \sin ^2(180-18)^{\circ} \\ =1+2 \sin ^2…
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