If {tan ^{ - 1}}x + {cos ^{ - 1}}frac{y}{{sqr | vedclass

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

Question

(A) Given equation: ${	an ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }}$We know that ${\cos ^{ - 1

Vedclass · Accepted Answer

$(1, 2)$ and $(2, 7)$

Vedclass · Answer

(A) Given equation: ${	an ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }}$We know that ${\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {	an ^{ - 1}}\frac{1}{y}$ and ${\sin ^{ - 1}}\frac{3}{{\sqrt {10} }} = {	an ^{ - 1}}3$.Substituting these,we get: ${	an ^{ - 1}}x + {	an ^{ - 1}}\frac{1}{y} = {	an ^{ - 1}}3$.Using the formula ${	an ^{ - 1}}A + {	an ^{ - 1}}B = {	an ^{ - 1}}\frac{A+B}{1-AB}$,we have: $\frac{x + \frac{1}{y}}{1 - \frac{x}{y}} = 3$.Simplifying: $\frac{xy + 1}{y - x} = 3 \implies xy + 1 = 3y - 3x \implies xy + 3x - 3y + 1 = 0$.Adding $8$ to both sides to factorize: $x(y + 3) - 3(y + 3) + 10 = 0 \implies (x - 3)(y + 3) = -10 \implies (3 - x)(y + 3) = 10$.Since $x, y$ are positive integers,we test factors of $10$: $(3-x)$ can be $1$ or $2$.If $3 - x = 1 \implies x = 2$,then $y + 3 = 10 \implies y = 7$.If $3 - x = 2 \implies x = 1$,then $y + 3 = 5 \implies y = 2$.Thus,the pairs are $(1, 2)$ and $(2, 7)$.

If ${\tan ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }}$ and both $x$ and $y$ are positive integers,then the possible values of $(x, y)$ are:

Similar Questions

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

Solve $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2},$ then $x$ is equal to

If $y = \tan^{-1} \left( \frac{\log(e/x^3)}{\log(ex^3)} \right) + \tan^{-1} \left( \frac{\log(e^4x^3)}{\log(e/x^{12})} \right)$, for $x \in (e^{-1/3}, e^{1/12})$, then $\frac{dy}{dx}$ is equal to...

The value of $\cos ^{-1} \left[ \cot \left( \sin ^{-1} \sqrt{\frac{2-\sqrt{3}}{4}} \right) + \cos ^{-1} \left( \frac{\sqrt{12}}{4} \right) + \sec ^{-1} \sqrt{2} \right]$ is

If $y = \tan ^{-1}\left(\frac{1}{1+x+x^{2}}\right) + \tan ^{-1}\left(\frac{1}{x^{2}+2x+3}\right) + \tan ^{-1}\left(\frac{1}{x^{2}+5x+7}\right) + \dots + n \text{ terms}$,then $y'(0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module