Show that $\tan ^{4} \theta+\tan ^{2} \theta=\sec ^{4} \theta-\sec ^{2} \theta$
(N/A) $L$.$H$.$S$. $= \tan ^{4} \theta + \tan ^{2} \theta$
$= \tan ^{2} \theta (\tan ^{2} \theta + 1)$
$= \tan ^{2} \theta \cdot \sec ^{2} \theta$ (Since $\sec ^{2} \theta = \tan ^{2} \theta + 1$)
$= (\sec ^{2} \theta - 1) \cdot \sec ^{2} \theta$ (Since $\tan ^{2} \theta = \sec ^{2} \theta - 1$)
$= \sec ^{4} \theta - \sec ^{2} \theta = \text{R.H.S.}$
$= \tan ^{2} \theta (\tan ^{2} \theta + 1)$
$= \tan ^{2} \theta \cdot \sec ^{2} \theta$ (Since $\sec ^{2} \theta = \tan ^{2} \theta + 1$)
$= (\sec ^{2} \theta - 1) \cdot \sec ^{2} \theta$ (Since $\tan ^{2} \theta = \sec ^{2} \theta - 1$)
$= \sec ^{4} \theta - \sec ^{2} \theta = \text{R.H.S.}$
Explore More
Similar Questions
If $\sec 4A = \operatorname{cosec}(A - 20^{\circ})$,where $4A$ is an acute angle,find the value of $A$ (in $^{\circ}$).
Easy
View Solution$\frac{1+\tan ^{2} A}{1+\cot ^{2} A} = \dots$
Easy
View Solution$\tan ^{2} \theta - \sec ^{2} \theta = \ldots \ldots \ldots$
Easy
View SolutionProve that $\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta},$ using the identity $\sec ^{2} \theta=1+\tan ^{2} \theta.$
Difficult
View SolutionProve that,$(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)=\sec \alpha+\operatorname{cosec} \alpha$
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo