Question

prove that tan ^2 - sin ^2 = tan ^2 sin ^2

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Answer 1

Step-by-step explanation:

To prove the trigonometric identity :

$\implies \sf tan^2 \theta - sin^2 \theta= tan^2 \theta sin ^2 \theta$

Let's take LHS :

$\implies \sf tan^2 \theta - sin^2 \theta = \dfrac{sin^2 \theta}{cos^2 \theta} - sin^2 \theta \\\\\\\bold{\boxed{\bold{tan^2\theta = \dfrac{sin^2 \theta}{cos^2 \theta}}}}$

$\implies \sf \bigg( \dfrac{sin \theta}{cos \theta} \bigg)^2 - \bigg(\dfrac{sin^2 \theta. cos^2 \theta}{cos^2 \theta} \bigg)$

$\implies \sf \dfrac{sin^2 \theta (1-cos^2 \theta)}{cos^2 \theta}$

$\implies \sf tan^2 \theta (1-cos^2 \theta)$

$\boxed{\bold{Formula: sin^2 \theta + cos^2 \theta = 1 \implies 1-cos^2 \theta = sin^2 \theta}}\\\\\implies \sf \bold{tan^2 \theta. sin^2 \theta}$

Hence proved

(A solution has also been attached.)