If y(alpha)=sqrt{2left(frac{tan alpha+cot alp | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $y(\alpha)=\sqrt{2\left(\frac{	an \alpha+\cot \alpha}{\sec^2 \alpha}ight)+\csc^2 \alpha}$.Since $1+	an^2 \alpha = \sec^2 \alpha$,we have

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given $y(\alpha)=\sqrt{2\left(\frac{	an \alpha+\cot \alpha}{\sec^2 \alpha}ight)+\csc^2 \alpha}$.Since $1+	an^2 \alpha = \sec^2 \alpha$,we have $y(\alpha)=\sqrt{2\left(\frac{\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \alpha}{\sin \alpha}}{\frac{1}{\cos^2 \alpha}}ight)+\csc^2 \alpha}$.Simplifying the expression inside the square root: $2\left(\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}ight) \cdot \cos^2 \alpha + \csc^2 \alpha = 2\frac{\cos \alpha}{\sin \alpha} + \csc^2 \alpha = 2\cot \alpha + \csc^2 \alpha$.Note that $2\cot \alpha + \csc^2 \alpha = 2\cot \alpha + 1 + \cot^2 \alpha = (1+\cot \alpha)^2$.Thus,$y(\alpha) = \sqrt{(1+\cot \alpha)^2} = |1+\cot \alpha|$.For $\alpha \in \left(\frac{3\pi}{4}, \piight)$,$\cot \alpha Therefore,$y(\alpha) = -(1+\cot \alpha) = -1 - \cot \alpha$.Differentiating with respect to $\alpha$: $\frac{dy}{d\alpha} = -(-\csc^2 \alpha) = \csc^2 \alpha$.At $\alpha = \frac{5\pi}{6}$,$\csc \alpha = \csc \frac{5\pi}{6} = 2$.So,$\frac{dy}{d\alpha} = (2)^2 = 4$.

If $y(\alpha)=\sqrt{2\left(\frac{\tan \alpha+\cot \alpha}{1+\tan ^{2} \alpha}\right)+\frac{1}{\sin ^{2} \alpha}}$ for $\alpha \in\left(\frac{3 \pi}{4}, \pi\right)$,then find $\frac{d y}{d \alpha}$ at $\alpha=\frac{5 \pi}{6}$.

Similar Questions

If $f(x) = e^x g(x)$,$g(0) = 2$,and $g'(0) = 1$,then $f'(0)$ is

If $f(x) = \tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$,then $\lim_{x \rightarrow \frac{1}{2}} \frac{2[f(x)-f(\frac{1}{2})]}{2x-1} = $

If $f(1)=1$ and $f^{\prime}(1)=3$,then the derivative of $f(f(f(x)))+(f(x))^2$ at $x=1$ is

If $f(x) = \sin \left(\cosh \left(\frac{x^2+1}{x^2+2}\right)\right)$, then $f^{\prime}(1) = $

If $y = \sin^{-1} \left( x\sqrt{1 - x} + \sqrt{x} \sqrt{1 - x^2} \right)$ and $\frac{dy}{dx} = \frac{1}{2\sqrt{x(1 - x)}} + p$,then $p =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module