TS EAMCET · Maths · Trigonometric Ratios & Identities
\(\cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\cos 76^{\circ} \cos 16^{\circ}\) is equal to
- A \(0\)
- B \(\frac{1}{2}\)
- C \(-\frac{1}{4}\)
- D \(\frac{3}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\cos 76^{\circ} \cos 16^{\circ} \\ & =\cos ^2 76^{\circ}+1-\sin ^2 16^{\circ}-\frac{1}{2} \cdot 2 \cos 76^{\circ} \cos 16^{\circ} \\ & =1+\cos \left(76^{\circ}+16^{\circ}\right) \cos \left(76^{\circ}-16^{\circ}\right) \\ &…
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