JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \(\left(\sin ^{-1} x\right)^{2}-\left(\cos ^{-1} x\right)^{2}=a ; 0\,<\,x\,<\,1, a \neq 0\), then the value of \(2 \mathrm{x}^{2}-1\) is :
- A \(\cos \left(\frac{4 \mathrm{a}}{\pi}\right)\)
- B \(\sin \left(\frac{2 \mathrm{a}}{\pi}\right)\)
- C \(\cos \left(\frac{2 \mathrm{a}}{\pi}\right)\)
- D \(\sin \left(\frac{4 \mathrm{a}}{\pi}\right)\)
Answer & Solution
Correct Answer
(B) \(\sin \left(\frac{2 \mathrm{a}}{\pi}\right)\)
Step-by-step Solution
Detailed explanation
\(\text { Given } \mathrm{a}=\left(\sin ^{-1} \mathrm{x}\right)^{2}-\left(\cos ^{-1} \mathrm{x}\right)^{2}\) \(=\left(\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}\right)\left(\sin ^{-1} \mathrm{x}-\cos ^{-1} \mathrm{x}\right)\)…
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