General Solution of Equation tanθ = tanα
General Solution of Equation: Given tanθ = tanα
\(\frac{\sin \theta }{\cos \theta }=\frac{\sin \alpha }{\cos \alpha }\),
sin θ cosα = sinα cosθ,
sin θ cosα – sinα cosθ = 0,
\(\left( \because \sin \left( A-B \right)=\sin A\cos B-\cos A\sin B \right)\),
sin (θ – α) = 0 ,
θ – α = Sin⁻¹(0),
\(\theta -\alpha =n\pi ,\ n\in Z\),
\(\theta =n\pi +\alpha ,\ n\in Z\),
Note: For general solution of the equation tanθ = k, where k ϵ R, We have tanθ = tan(tan⁻¹k). thus, \(\theta =n\pi +({{\tan }^{-1}}k),\ n\in Z\),
Example: Solve tan5θ = cot2θ
Solution: tan5θ = cot2θ
\(\tan 5\theta =\tan \left( \frac{\pi }{2}-2\theta \right)\),
\(5\theta =n\pi +\left( \frac{\pi}{2}-2\theta \right)\),
7θ =nπ + π/2,
θ =nπ/7 + π/14, where n ϵ Z, but n ≠ 3, 10, 17, . ……
where tan5θ is not defined.