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: $a \sin^2 x + b \cos^2 x = c$Divide by $\cos^2 x$: $a 	an^2 x + b = c \sec^2 x = c(1 + 	an^2 x)$$a 	an^2 x + b = c + c 	an^2 x \Rightar

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given: $a \sin^2 x + b \cos^2 x = c$Divide by $\cos^2 x$: $a 	an^2 x + b = c \sec^2 x = c(1 + 	an^2 x)$$a 	an^2 x + b = c + c 	an^2 x \Rightarrow (a - c) 	an^2 x = c - b \Rightarrow 	an^2 x = \frac{b - c}{a - c} = \frac{b - c}{-(c - a)} = \frac{b - c}{c - a}$Given: $b \sin^2 y + a \cos^2 y = d$Divide by $\cos^2 y$: $b 	an^2 y + a = d \sec^2 y = d(1 + 	an^2 y)$$b 	an^2 y + a = d + d 	an^2 y \Rightarrow (b - d) 	an^2 y = d - a \Rightarrow 	an^2 y = \frac{d - a}{b - d} = \frac{-(a - d)}{-(d - b)} = \frac{a - d}{d - b}$Given: $a 	an x = b 	an y \Rightarrow \frac{	an x}{	an y} = \frac{b}{a} \Rightarrow \frac{	an^2 x}{	an^2 y} = \frac{b^2}{a^2}$Substituting the values: $\frac{b^2}{a^2} = \frac{(b - c)/(c - a)}{(a - d)/(d - b)} = \frac{(b - c)(d - b)}{(c - a)(a - d)}$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

The value of $\tan 56^{\circ} - \tan 11^{\circ} - \tan 56^{\circ} \tan 11^{\circ}$ is

$\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} = $ (when $x$ lies in $II^{nd}$ quadrant)

If $\sin \theta_1 + \sin \theta_2 + \sin \theta_3 = 3$,then $\cos \theta_1 + \cos \theta_2 + \cos \theta_3 = $

If $\tan \alpha$ and $\tan \beta$ are the roots of the equation $x^2 + px + q = 0$ $(p \ne 0)$,then:

In a triangle $ABC$,if $A = \frac{\pi}{4}$ and $\tan B \tan C = K$,then $K$ must satisfy:

Vedclass Test Series

Exam Paper Generator

Online Exam Module