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Mathematics Question on Trigonometric Functions

If sin⁡(θ+α)=cos⁡(θ+α),\sin(\theta + \alpha) = \cos(\theta + \alpha), then the value of 1−tan⁡α1+tan⁡α\frac{1 - \tan \alpha}{1+ \tan \alpha} is

A

−tan⁡θ- \tan \theta

B

−cos⁡θ- \cos \theta

C

00

D

tan⁡θ\tan \theta

Answer

tan⁡θ\tan \theta

Explanation

Solution

sin⁡(θ+α)=cos⁡(θ+α),\sin(\theta + \alpha) = \cos(\theta + \alpha),
sin⁡θcos⁡α+cos⁡θsin⁡α=cos⁡θcos⁡α−sin⁡θsin⁡α\sin\theta \cos\alpha + \cos\theta \sin\alpha =\cos\theta \cos \alpha - \sin \theta \sin\alpha
sin⁡θ(cos⁡α+sin⁡α)=cos⁡θ(cos⁡α−sin⁡α)\sin \theta\left(\cos\alpha + \sin \alpha \right)= \cos \theta \left( \cos \alpha - \sin\alpha\right)
=tan⁡θ=cos⁡α−sin⁡αcos⁡α+sin⁡α=1−tan⁡α1+tan⁡α=\tan \theta= \frac{\cos\alpha- \sin \alpha }{\cos \alpha + \sin \alpha } = \frac{1 - \tan \alpha}{1+ \tan \alpha}