If a sin^2 x + b cos^2 x = c,b sin^2 y + a co | vedclass

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

Question

(B) Given $a \sin^2 x + b \cos^2 x = c$. Dividing by $\cos^2 x$,we get $a 	an^2 x + b = c \sec^2 x = c(1 + 	an^2 x)$.So,$(a - c) 	an^2 x = c - b$,w

Vedclass · Accepted Answer

$\frac{(a - d)(c - a)}{(b - c)(d - b)}$

Vedclass · Answer

(B) Given $a \sin^2 x + b \cos^2 x = c$. Dividing by $\cos^2 x$,we get $a 	an^2 x + b = c \sec^2 x = c(1 + 	an^2 x)$.So,$(a - c) 	an^2 x = c - b$,which implies $	an^2 x = \frac{b - c}{a - c} = \frac{c - b}{c - a}$.Similarly,for $b \sin^2 y + a \cos^2 y = d$,dividing by $\cos^2 y$ gives $b 	an^2 y + a = d(1 + 	an^2 y)$.So,$(b - d) 	an^2 y = d - a$,which implies $	an^2 y = \frac{d - a}{b - d} = \frac{a - d}{d - b}$.Given $a 	an x = b 	an y$,we have $\frac{	an x}{	an y} = \frac{b}{a}$.Squaring both sides,$\frac{	an^2 x}{	an^2 y} = \frac{b^2}{a^2}$.Substituting the values,$\frac{b^2}{a^2} = \frac{(c - b)/(c - a)}{(a - d)/(d - b)} = \frac{(b - c)(d - b)}{(c - a)(a - d)}$.Therefore,$\frac{a^2}{b^2} = \frac{(c - a)(a - d)}{(b - c)(d - b)}$.

If $a \sin^2 x + b \cos^2 x = c$,$b \sin^2 y + a \cos^2 y = d$ and $a \tan x = b \tan y$,then $\frac{a^2}{b^2}$ is equal to

Similar Questions

If $\sin (y+z-x), \sin (z+x-y)$ and $\sin (x+y-z)$ are in $A$.$P$.,then

If $2 \sinh^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=\log \left(\frac{1+x}{1-x}\right)$,then $x$ is equal to

If $\cos \theta - 4 \sin \theta = 1$,then $\sin \theta + 4 \cos \theta$ is equal to

$\frac{\cos 12^{\circ}-\sin 12^{\circ}}{\cos 12^{\circ}+\sin 12^{\circ}}+\frac{\sin 147^{\circ}}{\cos 147^{\circ}} = $

If $\cos \theta - \sin \theta = \sqrt{2} \sin \theta,$ then $\cos \theta + \sin \theta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module