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

A$\frac{{2{\rm{ab}}}}{{{{\rm{a}}^2} - {{\rm{b}}^2}}}$
B$\frac{{{{\rm{a}}^2} - {{\rm{b}}^2}}}{{2{\rm{ab}}}}$
C$\frac{{{\rm{ab}}}}{{{{\rm{a}}^2} - {{\rm{b}}^2}}}$
D$\frac{{{{\rm{a}}^2} - {{\rm{b}}^2}}}{{{\rm{ab}}}}$

Answer not set

Discussion 1

CCommunity09 Jul 2019

⇒ (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