Question

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Answer 1

Given : (a² - b²)(sinθ) + 2abcosθ = a² + b²

$:\implies$  a²sinθ - b²sinθ + 2abcosθ = a² + b²

$:\implies$  a²sinθ - a² - b²sinθ - b² + 2abcosθ = 0

$:\implies$  a²(sinθ - 1) - b²(sinθ + 1) + 2abcosθ = 0

$:\implies$  -a²(1 - sinθ) - b²(1 + sinθ) + 2abcosθ = 0

Multiplying the entire equation with -1, We get :

$:\implies$  a²(1 - sinθ) + b²(1 + sinθ) - 2abcosθ = 0

$\mathsf{\bigstar\;\;(1 - sin\theta) = \big(\sqrt{1 - sin\theta}\big)\big(\sqrt{1 - sin\theta}\big) = \big(\sqrt{1 - sin\theta}\big)^2}$

$\mathsf{\bigstar\;\;(1 + sin\theta) = \big(\sqrt{1 + sin\theta}\big)\big(\sqrt{1 + sin\theta}\big) = \big(\sqrt{1 + sin\theta}\big)^2}$

$\mathsf{\bigstar\;\;cos\theta = \sqrt{(1 - sin^2\theta)} = \big(\sqrt{1 - sin\theta}\big)\big(\sqrt{1 + sin\theta}\big)}$

$\mathsf{\implies a^2\big(\sqrt{1 - sin\theta}\big)^2 + b^2\big(\sqrt{1 + sin\theta}\big)^2 - 2ab\big(\sqrt{1 - sin\theta}\big)\big(\sqrt{1 + sin\theta}\big) = 0}$

$\mathsf{\implies \big(a\sqrt{1 - sin\theta}\big)^2 + \big(b\sqrt{1 + sin\theta}\big)^2 - 2ab\big(\sqrt{1 - sin\theta}\big)\big(\sqrt{1 + sin\theta}\big) = 0}$

This is in the form : a² + b² - 2ab which is equal to (a - b)²

$\mathsf{\implies \big(a\sqrt{1 - sin\theta} - b\sqrt{1 + sin\theta}\big)^2 = 0}$

$\mathsf{\implies a\sqrt{1 - sin\theta} - b\sqrt{1 + sin\theta} = 0}$

$\mathsf{\implies a\sqrt{1 - sin\theta} = b\sqrt{1 + sin\theta}}$

Squaring on both sides, We get :

$\mathsf{\implies a^2(1 - sin\theta) = b^2(1 + sin\theta)}$

$\mathsf{\implies a^2 - a^2sin\theta = b^2 + b^2sin\theta}$

$\mathsf{\implies a^2sin\theta + b^2sin\theta = a^2 - b^2}$

$\mathsf{\implies (sin\theta)(a^2 + b^2) = a^2 - b^2}$

$\mathsf{\implies sin\theta = \dfrac{a^2 - b^2}{a^2 + b^2}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{cos^2\theta = 1 - sin^2\theta}}}$

$\mathsf{\implies cos^2\theta = 1 - \dfrac{(a^2 - b^2)^2}{(a^2 + b^2)^2}}$

$\mathsf{\implies cos^2\theta = \dfrac{(a^2 + b^2)^2 - (a^2 - b^2)^2}{(a^2 + b^2)^2}}$

$\mathsf{\implies cos^2\theta = \dfrac{(a^4 + b^4 + 2a^2b^2 - a^4 - b^4 + 2a^2b^2)}{(a^2 + b^2)^2}}$

$\mathsf{\implies cos^2\theta = \dfrac{4a^2b^2}{(a^2 + b^2)^2}}$

$\mathsf{\implies cos^2\theta = \bigg(\dfrac{2ab}{a^2 + b^2}\bigg)^2}$

$\mathsf{\implies cos\theta = \dfrac{2ab}{a^2 + b^2}}$

$\bigstar\;\; \textsf{We know that : \boxed{\mathsf{tan\theta = \dfrac{sin\theta}{cos\theta}}}}$

$\mathsf{\implies tan\theta = \dfrac{\bigg(\dfrac{a^2 - b^2}{a^2 + b^2}\bigg)}{\bigg(\dfrac{2ab}{a^2 + b^2}\bigg)}}$

$\mathsf{\implies tan\theta = {\bigg(\dfrac{a^2 - b^2}{a^2 + b^2}\bigg)}{\bigg(\dfrac{a^2 + b^2}{2ab}\bigg)}}$

$\mathsf{\implies tan\theta = {\dfrac{a^2 - b^2}{2ab}}}$