If {1 – tan²(π/4 – x)}/{1 + tan²(π/4 – x)} = sin kx, Then Find k
Question:
\[ \frac{1-\tan^2\left(\frac{\pi}{4}-x\right)} {1+\tan^2\left(\frac{\pi}{4}-x\right)} = \sin kx \] Then find the value of \(k\).
Solution
Using the identity
\[ \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta \]Let
\[ \theta=\frac{\pi}{4}-x \]Therefore,
\[ \frac{1-\tan^2\left(\frac{\pi}{4}-x\right)} {1+\tan^2\left(\frac{\pi}{4}-x\right)} = \cos\left[2\left(\frac{\pi}{4}-x\right)\right] \] \[ =\cos\left(\frac{\pi}{2}-2x\right) \]Using the identity
\[ \cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta \]We get
\[ \cos\left(\frac{\pi}{2}-2x\right)=\sin 2x \]Given that
\[ \sin 2x=\sin kx \]Hence,
\[ k=2 \]