Question

If $\tan \theta =\dfrac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }$ , then show that $\sin \alpha +cos\alpha =\sqrt{2}\cos \theta $ .

Answer 1

Hint: First, start by simplifying the equation $\tan \theta =\dfrac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }$ to get a relation between $\theta \text{ and }\alpha $ . Once you get the relation, substitute the value of $\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 $\tan \theta =\dfrac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }$ . So, simplifying this, we get
$\tan \theta =\dfrac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }$
$\Rightarrow \tan \theta =\dfrac{\cos \alpha \left( \dfrac{\sin \alpha }{\cos \alpha }-1 \right)}{\cos \alpha \left( \dfrac{\sin \alpha }{\cos \alpha }+1 \right)}$
We know that $\tan A=\dfrac{\sin A}{\cos A}$ .
$\tan \theta =\dfrac{\dfrac{\sin \alpha }{\cos \alpha }-1}{\dfrac{\sin \alpha }{\cos \alpha }+1}$
$\Rightarrow \tan \theta =\dfrac{\tan \alpha -1}{\tan \alpha +1}$
Now we know that $\tan \left( A-45{}^\circ \right)=\dfrac{\tan A-1}{\tan A+1}$ . Therefore, we can say that:
$\tan \theta =\tan \left( \alpha -45{}^\circ \right)$
From here, we can conclude that $\theta =\left( \alpha -45{}^\circ \right)$ .
Now starting with the right-hand side of the equation that we need to prove. If we put $\theta =\alpha -45{}^\circ $ in the expression, we get:
$\sqrt{2}\cos \theta $
$=\sqrt{2}\cos \left( \alpha -45{}^\circ \right)$
Now we know that $\cos \left( A-45{}^\circ \right)=\dfrac{\cos A}{\sqrt{2}}+\dfrac{\sin A}{\sqrt{2}}$ . On using this in our expression, we get
$=\sqrt{2}\left( \dfrac{\cos \alpha }{\sqrt{2}}+\dfrac{\sin \alpha }{\sqrt{2}} \right)$
$=\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 $\tan \theta =\tan \left( \alpha -45{}^\circ \right)$ actually implies that $\theta =n\pi +\left( \alpha -45{}^\circ \right)$, and $\theta =\alpha -45{}^\circ $ is just one of the many possible solutions that we used in the above question.