સાબિત કરો કે $\tan ^{4} \theta+\tan ^{2} \theta=\sec ^{4} \theta-\sec ^{2} \theta$
(N/A) ડા.બા. ($L$.$H$.$S$.) $= \tan ^{4} \theta + \tan ^{2} \theta$
$= \tan ^{2} \theta (\tan ^{2} \theta + 1)$
$= \tan ^{2} \theta \cdot \sec ^{2} \theta$ (કારણ કે $\sec ^{2} \theta = \tan ^{2} \theta + 1$)
$= (\sec ^{2} \theta - 1) \cdot \sec ^{2} \theta$ (કારણ કે $\tan ^{2} \theta = \sec ^{2} \theta - 1$)
$= \sec ^{4} \theta - \sec ^{2} \theta = \text{જ.બા. (R.H.S.)}$
$= \tan ^{2} \theta (\tan ^{2} \theta + 1)$
$= \tan ^{2} \theta \cdot \sec ^{2} \theta$ (કારણ કે $\sec ^{2} \theta = \tan ^{2} \theta + 1$)
$= (\sec ^{2} \theta - 1) \cdot \sec ^{2} \theta$ (કારણ કે $\tan ^{2} \theta = \sec ^{2} \theta - 1$)
$= \sec ^{4} \theta - \sec ^{2} \theta = \text{જ.બા. (R.H.S.)}$
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