Question

[1+tan^2theta]×sin^2theta/tantheta=tantheta

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Answer 1

Given: $\dfrac{(1 + tan^{2} \theta) \times sin^2 \theta}{tan \theta} = tan \theta$

Proof:

LHS = $\dfrac{(1 + tan^{2} \theta) \times sin^2 \theta}{tan \theta}$

⇒ $\dfrac{(1 + \dfrac{sin^2\theta}{cos^2\theta})\times sin^2\theta }{tan\theta} \\\\\dfrac{(\dfrac{cos^2\theta + sin^2\theta}{cos^2\theta})\times sin^2\theta }{tan\theta}\\\\\dfrac{(\dfrac{1}{cos^2\theta})\times sin^2\theta }{tan\theta}\\\\\dfrac{(\dfrac{sin^2\theta}{cos^2\theta})}{tan\theta}\\\\\dfrac{tan^2\theta}{tan\theta}\\\\tan\theta$

= RHS

Hence Proved.

Trigonometric Identities used:

• sin²Θ + cos²Θ =1

• tan²Θ = $\dfrac{sin^2\theta}{cos^2\theta}$