Question

Prove that tan²θ−sin²θ= sin⁴θsec²θ

Answer 1

Step-by-step explanation:

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Answer 2

$\bigstar \:\: \underline{\underline{\bf \red{To\:Prove\: :}}}$

$\longmapsto \tan^2 \theta -\sin^2 \theta= \sin^4 \theta\sec^2 \theta$

$\bigstar \:\: \underline{\underline{\bf \red{Solution\: :}}}$

$\blue \longmapsto \sf\tan^2\theta-\sin^2 = \sin^4\theta.\sec^2\theta$

$\large\bf L.H.S$

$\blue \longmapsto \tan^2 \theta-\sin^2 \theta$

$\red\bigstar\;\;\boxed{tan\theta=\frac{sin\theta}{\cos\theta}}$

$\implies \sf\dfrac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta\\\\\\\implies\dfrac{\sin^2\theta-\sin^2\theta \cos^2\theta}{\cos^2\theta}\\\\\\\implies\dfrac{sin^2\theta(1- \cos^2\theta)}{\cos^2\theta}\\\\\red\bigstar\;\;\boxed{\sf\sin^2\theta=1-\cos^2\theta}\\\\\\\implies\dfrac{sin^2\theta(\sin^2\theta)}{\cos^2\theta}\\\\\\\implies \dfrac{\sin^4\theta}{\cos^2\theta} \\\\\\\implies \sin^4\theta.\dfrac{1}{\cos^2\theta}\\\\\\\red\bigstar\boxed{\sec\theta=\frac{1}{\cos\theta}}\\\\\\\implies \pink{\sin^4\theta\sec^2\theta }$

$\large\bf R.H.S$

$\;\;\;\;\;\;\;\;\;\;\; \large\huge\underline{\underline{\green{\bf \mathfrak{Hence\; proved}}}}$