If (a2 – b2) sin θ + 2abcosθ = a2 + b2, then tan θ

If (a2 – b2) sin θ + 2abcosθ = a2 + b2, then tan θ
1). $\frac{{2{\rm{ab}}}}{{{{\rm{a}}^2} - {{\rm{b}}^2}}}$
2). $\frac{{{{\rm{a}}^2} - {{\rm{b}}^2}}}{{2{\rm{ab}}}}$
3). $\frac{{{\rm{ab}}}}{{{{\rm{a}}^2} - {{\rm{b}}^2}}}$
4). $\frac{{{{\rm{a}}^2} - {{\rm{b}}^2}}}{{{\rm{ab}}}}$

⇒ (a² – b²) sin θ + 2abcos θ = a² + b²

⇒ Divide both sides by (a² + b²)

⇒ [(a² - b²)/(a² + b²)] sin θ + [2ab/(a² + b²)] cos θ = 1

⇒ Sin θ = [(a² - b²)/(a² + b²)]        ---- 1

⇒ cos θ = [2ab/(a² + b²)]        ---- 2

⇒ From equation 1 and 2

∴ tan θ = (a² - b²)/2ab

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