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If (a2 - b2) sinx + ab cosx = a2 + b2 then find the value of cosθ.tanθ |
\(\frac{a^2-b^2}{2ab}\) \(\frac{a^2+b^2}{a^2-b^2}\) \(\frac{a^2-b^2}{a^2+b^2}\) \(\frac{a+b}{a^2-b^2}\) |
\(\frac{a^2-b^2}{a^2+b^2}\) |
Make sin2θ + cos2θ = 1 form so divide the equation by a2 + b2 \(\frac{a^2-b^2}{a^2+b^2}\)sinx + \(\frac{2ab}{a^2+b^2}\)cosx = 1
P = a2 - b2 H = a2 + b2 B = 2ab ⇒ cosx.tanx = \(\left(\frac{2ab}{a^2+b^2} \right) \left(\frac{a^2-b^2}{2ab} \right)\) = \(\frac{a^2-b^2}{a^2+b^2}\) |
