ત્રિકોણમિતિય વિધયો — ગણિત ધોરણ 11 સાયન્સ — Question

$\frac{sec^4 \theta}{a} + \frac{tan^4 \theta}{b} = \frac{1}{a + b}$

$\therefore \frac{sec^4 \theta}{a} + \frac{tan^4 \theta}{b} = \frac{sec^2 \theta - tan^2 \theta}{a + b}$

$\therefore \frac{sec^2 \theta}{a (a + b)} \left[\left(a + b\right) sec^2 \theta - a\right]$$ + \frac{tan^2 \theta}{(a + b) b} \left[\left(a + b\right) tan^2 \theta + b\right] = 0$

$\therefore a tan^2 \theta + b sec^2 \theta = 0$

$\therefore sin^2 \theta = -\frac{b}{a} \leq 1$

$|\frac{b}{a}| \leq 1 \Rightarrow |b| \leq |a|$