JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
\(\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right)\) is equal to:
- A 1
- B 0
- C \(\frac{32}{65}\)
- D \(\frac{33}{65}\)
Answer & Solution
Correct Answer
(B) 0
Step-by-step Solution
Detailed explanation
\begin{aligned} & \cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right) \\ & \cos \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{33}{56}\right) \\ & \cos \left(\tan…
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