Mathematics - Area Under Curves Question with Solution | TestHub

$y=\sin x$ 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 $=\theta ={\tan }^{-1}\left\{\frac{\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)}{1+\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right)}\right\}={\tan }^{-1}\left(\frac{\sqrt{2}}{\frac{1}{2}}\right)$ $={\tan }^{-1}(2\sqrt{2})$