MHT CET · Maths · Area Under Curves
The angle between the curves \(y=\sin x\) and \(y=\cos x 0 - A \(\tan ^{-1}(\sqrt{2})\)
- B \(\tan ^{-1}(3 \sqrt{2})\)
- C \(\tan ^{-1}(2 \sqrt{2})\)
- D \(\tan ^{-1}(3 \sqrt{3})\)
- A \(\tan ^{-1}(\sqrt{2})\)
- B \(\tan ^{-1}(3 \sqrt{2})\)
- C \(\tan ^{-1}(2 \sqrt{2})\)
- D \(\tan ^{-1}(3 \sqrt{3})\)
Answer & Solution
Correct Answer
(C) \(\tan ^{-1}(2 \sqrt{2})\)
Step-by-step Solution
Detailed explanation
\(y=\sin x\) and \(y=\cos x\) intersects at \(x=\frac{\pi}{4}\)
\(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\text {for first curve at } x=\frac{\pi}{4}}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\)
\(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\text {for second curve at } x=\frac{\pi}{4}}=-\sin \frac{\pi}{4}=-\frac{1}{\sqrt{2}}\)
Required angle \(=\theta=\tan ^{-1}\left\{\frac{\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)}{1+\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right)}\right\}=\tan ^{-1}\left(\frac{\sqrt{2}}{\frac{1}{2}}\right)\)
\(=\tan ^{-1}(2 \sqrt{2})\)
\(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\text {for first curve at } x=\frac{\pi}{4}}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\)
\(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\text {for second curve at } x=\frac{\pi}{4}}=-\sin \frac{\pi}{4}=-\frac{1}{\sqrt{2}}\)
Required angle \(=\theta=\tan ^{-1}\left\{\frac{\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)}{1+\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right)}\right\}=\tan ^{-1}\left(\frac{\sqrt{2}}{\frac{1}{2}}\right)\)
\(=\tan ^{-1}(2 \sqrt{2})\)
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