MHT CET · Maths · Trigonometric Ratios & Identities
If \(\mathrm{A}+\mathrm{B}=\frac{\pi}{2}\) then the maximum value of \(\cos \mathrm{A} \cdot \cos \mathrm{B}\) is
- A \(\frac{1}{\sqrt{2}}\)
- B \(\frac{1}{2}\)
- C \(-\frac{1}{2}\)
- D \(-\frac{1}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{B}=\frac{\pi}{2}-\mathrm{A}\) \(\cos \mathrm{A} \cdot \cos \mathrm{B} = \cos \mathrm{A} \cdot \cos \left(\frac{\pi}{2}-\mathrm{A}\right)\)
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