Question

If $\tan \theta =\frac{\sin \left(\alpha \right)-\cos \left(\alpha \right)}{\sin \left(\alpha \right)+\cos \left(\alpha \right)}$, then $\sin \left(\alpha \right)+\cos \left(\alpha \right)$ and $\sin \left(\alpha \right)-\cos \left(\alpha \right)$ must be equal to

• A. $\sqrt{2}\cos \left(\theta \right),\sqrt{2}\sin \left(\theta \right)$
• B. $\sqrt{2}\sin \left(\theta \right),\sqrt{2}\cos \left(\theta \right)$
• C. $\sqrt{2}\sin \left(\theta \right),\sqrt{2}\sin \left(\theta \right)$
• D. $\sqrt{2}\cos \left(\theta \right),\sqrt{2}\cos \left(\theta \right)$

Answer 1

The correct option is A

$\sqrt{2}\cos \left(\theta \right),\sqrt{2}\sin \left(\theta \right)$

Explanation for the correct option:

Step 1. Simplify the equation

$\tan \theta =\frac{\sin \left(\alpha \right)-\cos \left(\alpha \right)}{\sin \left(\alpha \right)+\cos \left(\alpha \right)}$

dividing $\cos \left(\alpha \right)$ to both numerator and denominator

$$\begin{gathered} \tan \theta =\frac{{\displaystyle \frac{\sin \left(\alpha \right)}{\cos \left(\alpha \right)}}-{\displaystyle \frac{\cos \left(\alpha \right)}{\cos \left(\alpha \right)}}}{{\displaystyle \frac{\sin \left(\alpha \right)}{\cos \left(\alpha \right)}}+{\displaystyle \frac{\cos \left(\alpha \right)}{\cos \left(\alpha \right)}}} \\ \Rightarrow \tan \theta =\frac{{\displaystyle \tan \left(\alpha \right)-1}}{{\displaystyle \tan \left(\alpha \right)+1}}\left[\because tan\left(x\right)=\frac{sin\left(x\right)}{cos\left(x\right)}\right] \\ \Rightarrow \tan \theta =\frac{{\displaystyle \tan \left(\alpha \right)-\tan \left(\frac{\mathrm{\pi }}{4}\right)}}{{\displaystyle 1+\tan \left(\alpha \right)\times \tan \left(\frac{\mathrm{\pi }}{4}\right)}}\left[\because tan\left(\frac{\pi }{4}\right)=1\right] \\ \Rightarrow \tan \theta =\tan \left(\alpha -\frac{\mathrm{\pi }}{4}\right)\left[\because tan(x-y)=tan\left(\frac{x-y}{1+xy}\right)\right] \end{gathered}$$

Comparing both sides

$$\begin{gathered} \theta =\alpha -\frac{\mathrm{\pi }}{4} \\ \Rightarrow \alpha =\theta +\frac{\pi }{4} \end{gathered}$$

Step 2. Find the value of $\sin \left(\alpha \right)-\cos \left(\alpha \right)$

$$\begin{gathered} \sin \left(\alpha \right)-\cos \left(\alpha \right)=\sin \left(\theta +\frac{\mathrm{\pi }}{4}\right)-\cos \left(\theta +\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \sin \left(\alpha \right)-\cos \left(\alpha \right)=\cos \left(\theta \right)\sin \left(\frac{\mathrm{\pi }}{4}\right)+\sin \left(\theta \right)\cos \left(\frac{\mathrm{\pi }}{4}\right)-\left[\cos \left(\theta \right)\cos \left(\frac{\mathrm{\pi }}{4}\right)-\sin \left(\theta \right)\sin \left(\frac{\mathrm{\pi }}{4}\right)\right]\mathbf{}\left[\because sin(x+y)=sin\left(x\right)cos\left(y\right)+cos\left(y\right)sin\left(x\right),cos(x+y)=cos\left(x\right)cos\left(y\right)-sin\left(x\right)sin\left(y\right)\right] \\ \Rightarrow \sin \left(\alpha \right)-\cos \left(\alpha \right)=\cos \left(\theta \right)\frac{1}{\sqrt{2}}+\sin \left(\theta \right)\frac{1}{\sqrt{2}}-\left[\cos \left(\theta \right)\frac{1}{\sqrt{2}}-\sin \left(\theta \right)\frac{1}{\sqrt{2}}\right] \\ \Rightarrow \sin \left(\alpha \right)-\cos \left(\alpha \right)=\frac{1}{\sqrt{2}}\left[\cos \left(\theta \right)+\sin \left(\theta \right)-\cos \left(\theta \right)+\sin \left(\theta \right)\right] \\ \Rightarrow \sin \left(\alpha \right)-\cos \left(\alpha \right)=\frac{1}{\sqrt{2}}\left[2\sin \left(\theta \right)\right] \\ \Rightarrow \sin \left(\alpha \right)-\cos \left(\alpha \right)=\sqrt{2}\sin \left(\theta \right) \end{gathered}$$

Step 3. Find the value of $\sin \left(\alpha \right)+\cos \left(\alpha \right)$

$$\begin{gathered} \sin \left(\alpha \right)+\cos \left(\alpha \right)=\sin \left(\theta +\frac{\mathrm{\pi }}{4}\right)+\cos \left(\theta +\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \sin \left(\alpha \right)+\cos \left(\alpha \right)=\cos \left(\theta \right)\sin \left(\frac{\mathrm{\pi }}{4}\right)+\sin \left(\theta \right)\cos \left(\frac{\mathrm{\pi }}{4}\right)+\left[\cos \left(\theta \right)\cos \left(\frac{\mathrm{\pi }}{4}\right)-\sin \left(\theta \right)\sin \left(\frac{\mathrm{\pi }}{4}\right)\right]\mathbf{}\left[\because sin(x+y)=sin\left(x\right)cos\left(y\right)+cos\left(y\right)sin\left(x\right),cos(x+y)=cos\left(x\right)cos\left(y\right)-sin\left(x\right)sin\left(y\right)\right] \\ \Rightarrow \sin \left(\alpha \right)+\cos \left(\alpha \right)=\cos \left(\theta \right)\frac{1}{\sqrt{2}}+\sin \left(\theta \right)\frac{1}{\sqrt{2}}+\left[\cos \left(\theta \right)\frac{1}{\sqrt{2}}-\sin \left(\theta \right)\frac{1}{\sqrt{2}}\right] \\ \Rightarrow \sin \left(\alpha \right)+\cos \left(\alpha \right)=\frac{1}{\sqrt{2}}\left[\cos \left(\theta \right)+\sin \left(\theta \right)+\cos \left(\theta \right)-\sin \left(\theta \right)\right] \\ \Rightarrow \sin \left(\alpha \right)+\cos \left(\alpha \right)=\frac{1}{\sqrt{2}}\left[2\cos \left(\theta \right)\right] \\ \Rightarrow \sin \left(\alpha \right)+\cos \left(\alpha \right)=\sqrt{2}\cos \left(\theta \right) \end{gathered}$$Hence, option A is correct.