First slide

Trigonometric Identities

Question

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

Moderate

A

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

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

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

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

Solution

We have
$\begin{array}{l} a \sin^{2} x + b \cos^{2} x = c \\ \Rightarrow a \tan^{2} x + b = {csec}^{2} x \\ \Rightarrow \tan^{2} x = \frac{c - b}{a - c} \\ s i m i l a r l y \tan^{2} y = \frac{d - a}{b - d} \\ ∴ \frac{a^{2}}{b^{2}} = \frac{\tan^{2} y}{\tan^{2} x} = \frac{(a - d) (c - a)}{(b - c) (d - b)} \end{array}$

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