If sin ^{-1} x+sin ^{-1} y=frac{pi}{3} and co | vedclass

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

Question

(C) Given that $\cot ^{-1}\left(\frac{1}{x}\right)-\cot ^{-1}\left(\frac{1}{y}\right)=0$, which implies $\cot ^{-1}\left(\frac{1}{x}\right)=\cot ^{-1}

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) Given that $\cot ^{-1}\left(\frac{1}{x}\right)-\cot ^{-1}\left(\frac{1}{y}\right)=0$, which implies $\cot ^{-1}\left(\frac{1}{x}\right)=\cot ^{-1}\left(\frac{1}{y}\right)$.This means $\frac{1}{x}=\frac{1}{y}$, so $x=y$.Substitute $x=y$ into the first equation: $\sin ^{-1} x+\sin ^{-1} x=\frac{\pi}{3}$.$2 \sin ^{-1} x=\frac{\pi}{3} \implies \sin ^{-1} x=\frac{\pi}{6}$.Therefore, $x=\sin \left(\frac{\pi}{6}\right)=\frac{1}{2}$.Since $x=y$, we have $y=\frac{1}{2}$.Now, calculate $2 x^2+y^2-x y$:$2 \left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = 2 \left(\frac{1}{4}\right)+\frac{1}{4}-\frac{1}{4} = \frac{1}{2}+0 = \frac{1}{2}$.

If $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{3}$ and $\cot ^{-1}\left(\frac{1}{x}\right)-\cot ^{-1}\left(\frac{1}{y}\right)=0$, then $2 x^2+y^2-x y=$

Similar Questions

If $\alpha$ and $\beta$ are angles in the first quadrant such that $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{\sqrt{10}}$,then $\alpha + 2\beta =$ (in $^{\circ}$)

The value of $\cot^{-1} \left[ \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} \right]$, where $x \in (0, \frac{\pi}{2})$ is...

$\begin{aligned} & \text{If } \cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\} \\ & b>a, \text{ then } x= \end{aligned}$

In a $\triangle ABC$,if $\angle A = 90^{\circ}$,then $\cos^{-1}\left(\frac{R}{r_2+r_3}\right)$ is equal to (in $^{\circ}$)

If $(\sin ^{-1} x)^{2}-(\cos ^{-1} x)^{2}=a ; 0 < x < 1, a \neq 0$,then the value of $2 x^{2}-1$ is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module