Let x, y be real numbers such that x neq y an | vedclass

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

Question

(D) We know that $\sec(	an^{-1} x) = \sqrt{1+x^2}$. Given the equations $ax + b\sqrt{1+x^2} = c$ and $ay + b\sqrt{1+y^2} = c$,we can rewrite them as

Vedclass · Accepted Answer

$\frac{2ac}{a^2-c^2}$

Vedclass · Answer

(D) We know that $\sec(	an^{-1} x) = \sqrt{1+x^2}$. Given the equations $ax + b\sqrt{1+x^2} = c$ and $ay + b\sqrt{1+y^2} = c$,we can rewrite them as $b\sqrt{1+x^2} = c - ax$ and $b\sqrt{1+y^2} = c - ay$.Squaring both sides,we get $b^2(1+x^2) = c^2 - 2acx + a^2x^2$ and $b^2(1+y^2) = c^2 - 2acy + a^2y^2$.Rearranging these,we have $(a^2-b^2)x^2 - 2acx + (c^2-b^2) = 0$ and $(a^2-b^2)y^2 - 2acy + (c^2-b^2) = 0$.This implies that $x$ and $y$ are the distinct roots of the quadratic equation $(a^2-b^2)t^2 - 2act + (c^2-b^2) = 0$.Using the properties of roots,the sum of roots $x+y = \frac{2ac}{a^2-b^2}$ and the product of roots $xy = \frac{c^2-b^2}{a^2-b^2}$.Now,calculate $1-xy = 1 - \frac{c^2-b^2}{a^2-b^2} = \frac{a^2-b^2-c^2+b^2}{a^2-b^2} = \frac{a^2-c^2}{a^2-b^2}$.Therefore,$\frac{x+y}{1-xy} = \frac{2ac / (a^2-b^2)}{(a^2-c^2) / (a^2-b^2)} = \frac{2ac}{a^2-c^2}$.

Let $x, y$ be real numbers such that $x \neq y$ and $xy \neq 1$. If $ax + b \sec(\tan^{-1} x) = c$ and $ay + b \sec(\tan^{-1} y) = c$,then $\frac{x+y}{1-xy} =$

Similar Questions

If $p$ and $q$ are the roots of $x^2 + px + q = 0$,then

Let $\alpha, \beta$ be the roots of the equation $x^2-ax-b=0$ with $\operatorname{Im}(\alpha) < \operatorname{Im}(\beta)$. Let $P_n=\alpha^n-\beta^n$. If $P_3=-5 \sqrt{7} i, P_4=-3 \sqrt{7} i, P_5=11 \sqrt{7} i$ and $P_6=45 \sqrt{7} i$,then $|\alpha^4+\beta^4|$ is equal to . . . . . .

The roots of the equation $x^2 + ax + b = 0$ are $p$ and $q$. Then the equation whose roots are $p^2q$ and $pq^2$ will be:

The value of $a$ for which the sum of the squares of the roots of the equation $x^2-(a-2)x-(a+1)=0$ assumes the least value is

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+px^2+qx+r=0$,then $\alpha^3+\beta^3+\gamma^3=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module