If $\beta$ is an angle between the normals drawn to the curve $x^2+3 y^2=9$ at the points $(3 \cos \theta, \sqrt{3} \sin \theta)$ and $(-3 \sin \theta, \sqrt{3} \cos \theta), \theta \in\left(0, \frac{\pi}{2}\right)$, then
$\tan \beta=\frac{1}{\sqrt{3}} \sec 2 \theta$
$\cot \beta=\sqrt{3} \operatorname{cosec} 2 \theta$
$\sqrt{3} \cot \beta=\sin 2 \theta$
$\cot \beta=\frac{1}{\sqrt{2}} \sec 2 \theta$
If the area of a right-angle triangle with hypotenuse 5 is maximum, then its perimeter is
12
$2 \sqrt{3}+\sqrt{13}+5$
$7+\sqrt{21}$
$5(\sqrt{2}+1)$
$$ \int\left(\sum_{r=0}^{\infty} \frac{x^r 2^r}{r!}\right) d x= $$
$e^x+C$
$\frac{-2}{1-2 x}+C$
$2 e^{2 x}+C$
$\frac{e^{2 x}}{2}+C$
$$ \int \frac{d x}{12 \cos x+5 \sin x}= $$
$\frac{1}{13} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}-\frac{1}{2} \tan ^{-1} \frac{5}{12}\right)\right|+C$
$\frac{5}{12} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}-\frac{1}{2} \tan ^{-1} \frac{5}{12}\right)\right|+C$
$\frac{1}{13} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}+\frac{1}{2} \tan ^{-1} \frac{5}{12}\right)\right|+C$
$\frac{5}{12} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}+\frac{1}{2} \tan ^{-1} \frac{5}{12}\right)\right|+C$
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