(tan ^{-1} x)^2+(cot ^{-1} x)^2=frac{5 pi^2}{ | vedclass

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

Question

(A) Given equation is $(	an ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$.Using the identity $(	an ^{-1} x + \cot ^{-1} x)^2 = (	an ^{-1} x)^2 + (

Vedclass · Accepted Answer

-$1$

Vedclass · Answer

(A) Given equation is $(	an ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$.Using the identity $(	an ^{-1} x + \cot ^{-1} x)^2 = (	an ^{-1} x)^2 + (\cot ^{-1} x)^2 + 2 	an ^{-1} x \cot ^{-1} x$,we have:$(	an ^{-1} x + \cot ^{-1} x)^2 - 2 	an ^{-1} x \cot ^{-1} x = \frac{5 \pi^2}{8}$.Since $	an ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$,we substitute this into the equation:$(\frac{\pi}{2})^2 - 2 	an ^{-1} x (\frac{\pi}{2} - 	an ^{-1} x) = \frac{5 \pi^2}{8}$.$\frac{\pi^2}{4} - \pi 	an ^{-1} x + 2(	an ^{-1} x)^2 = \frac{5 \pi^2}{8}$.$2(	an ^{-1} x)^2 - \pi 	an ^{-1} x + \frac{\pi^2}{4} - \frac{5 \pi^2}{8} = 0$.$2(	an ^{-1} x)^2 - \pi 	an ^{-1} x - \frac{3 \pi^2}{8} = 0$.Let $u = 	an ^{-1} x$. Then $2u^2 - \pi u - \frac{3 \pi^2}{8} = 0$.Using the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$u = \frac{\pi \pm \sqrt{\pi^2 - 4(2)(-\frac{3 \pi^2}{8})}}{4} = \frac{\pi \pm \sqrt{\pi^2 + 3 \pi^2}}{4} = \frac{\pi \pm 2 \pi}{4}$.So,$u = \frac{3 \pi}{4}$ or $u = -\frac{\pi}{4}$.Since $x = 	an(u)$,we have $x = 	an(\frac{3 \pi}{4}) = -1$ or $x = 	an(-\frac{\pi}{4}) = -1$.Thus,$x = -1$.

$(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8} \Rightarrow x=$

Similar Questions

Whale is

$x$ grams of calcium carbonate was completely burnt in air. The weight of the solid residue formed is $28 \text{ g}$. What is the value of $x$ (in grams)?

If $A$ is the areal velocity of a planet of mass $M$,its angular momentum is

The combination of the gates shown in the figure below produces which gate?

An $X$-ray tube produces a continuous spectrum of radiation with its shortest wavelength of $45 \times 10^{-2} \text{ \AA}$. The maximum energy of a photon in the radiation in $eV$ is $(h = 6.62 \times 10^{-34} \text{ J-s}, c = 3 \times 10^8 \text{ m/s})$. (in $27,500$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module