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Question: If \(\tan \theta =\dfrac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }\) , then show that \(...

If tan⁡θ=sin⁡α−cos⁡αsin⁡α+cos⁡α\tan \theta =\dfrac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha } , then show that sin⁡α+cosα=2cos⁡θ\sin \alpha +cos\alpha =\sqrt{2}\cos \theta .

Explanation

Solution

Hint: First, start by simplifying the equation tan⁡θ=sin⁡α−cos⁡αsin⁡α+cos⁡α\tan \theta =\dfrac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha } to get a relation between θ and α\theta \text{ and }\alpha . Once you get the relation, substitute the value of θ\theta in the right-hand side of the equation that we need to prove. Solve the expression to get the answer.

Complete step-by-step answer:
It is given in the question that tan⁡θ=sin⁡α−cos⁡αsin⁡α+cos⁡α\tan \theta =\dfrac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha } . So, simplifying this, we get
tan⁡θ=sin⁡α−cos⁡αsin⁡α+cos⁡α\tan \theta =\dfrac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }
⇒tan⁡θ=cos⁡α(sin⁡αcos⁡α−1)cos⁡α(sin⁡αcos⁡α+1)\Rightarrow \tan \theta =\dfrac{\cos \alpha \left( \dfrac{\sin \alpha }{\cos \alpha }-1 \right)}{\cos \alpha \left( \dfrac{\sin \alpha }{\cos \alpha }+1 \right)}
We know that tan⁡A=sin⁡Acos⁡A\tan A=\dfrac{\sin A}{\cos A} .
tan⁡θ=sin⁡αcos⁡α−1sin⁡αcos⁡α+1\tan \theta =\dfrac{\dfrac{\sin \alpha }{\cos \alpha }-1}{\dfrac{\sin \alpha }{\cos \alpha }+1}
⇒tan⁡θ=tan⁡α−1tan⁡α+1\Rightarrow \tan \theta =\dfrac{\tan \alpha -1}{\tan \alpha +1}
Now we know that tan⁡(A−45∘)=tan⁡A−1tan⁡A+1\tan \left( A-45{}^\circ \right)=\dfrac{\tan A-1}{\tan A+1} . Therefore, we can say that:
tan⁡θ=tan⁡(α−45∘)\tan \theta =\tan \left( \alpha -45{}^\circ \right)
From here, we can conclude that θ=(α−45∘)\theta =\left( \alpha -45{}^\circ \right) .
Now starting with the right-hand side of the equation that we need to prove. If we put θ=α−45∘\theta =\alpha -45{}^\circ in the expression, we get:
2cos⁡θ\sqrt{2}\cos \theta
=2cos⁡(α−45∘)=\sqrt{2}\cos \left( \alpha -45{}^\circ \right)
Now we know that cos⁡(A−45∘)=cos⁡A2+sin⁡A2\cos \left( A-45{}^\circ \right)=\dfrac{\cos A}{\sqrt{2}}+\dfrac{\sin A}{\sqrt{2}} . On using this in our expression, we get
=2(cos⁡α2+sin⁡α2)=\sqrt{2}\left( \dfrac{\cos \alpha }{\sqrt{2}}+\dfrac{\sin \alpha }{\sqrt{2}} \right)
=cos⁡α+sin⁡α=\cos \alpha +\sin \alpha
The left-hand side of the equation given in the question is equal to the right-hand side of the equation. Hence, we can say that we have proved the equation given in the question.

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is 1+x-(x-1)=1+x-x-1. Also, be careful about the signs in the formula of sin(A-B) and sin(A+B). Also, you should know that tan⁡θ=tan⁡(α−45∘)\tan \theta =\tan \left( \alpha -45{}^\circ \right) actually implies that θ=nπ+(α−45∘)\theta =n\pi +\left( \alpha -45{}^\circ \right), and θ=α−45∘\theta =\alpha -45{}^\circ is just one of the many possible solutions that we used in the above question.