Question

If $\tan \left(\theta \right)+\sin \left(\theta \right)=m$ and $\tan \left(\theta \right)-\sin \left(\theta \right)=n$ then

• A. $m^2-n^2=4mn$
• B. $m^2-n^2=m^2+n^2$
• C. $m^2-n^2=4\sqrt{mn}$

Answer 1

The correct option is C

$m^2-n^2=4\sqrt{mn}$

Explanation for correct option

Given: $\tan \left(\theta \right)+\sin \left(\theta \right)=m$ and $\tan \left(\theta \right)-\sin \left(\theta \right)=n$

$$\begin{gathered} m^2-n^2={\left[\tan \left(\theta \right)-\sin \left(\theta \right)\right]}^2-{\left[\tan \left(\theta \right)+\sin \left(\theta \right)\right]}^2 \\ \Rightarrow (m+n)(m-n)=\left(\tan \left(\theta \right)-\sin \left(\theta \right)+\tan \left(\theta \right)+\sin \left(\theta \right)\right)\left(\tan \left(\theta \right)+\sin \left(\theta \right)-\tan \left(\theta \right)+\sin \left(\theta \right)\right)\left[\because a^2-b^2=(a-b)(a+b)\right] \\ \Rightarrow m^2-n^2=\left(2\tan \left(\theta \right)\right)\left(2\sin \left(\theta \right)\right) \\ \Rightarrow m^2-n^2=4\left(\tan \left(\theta \right)\right)\left(\sin \left(\theta \right)\right) \end{gathered}$$

Now,

$$\begin{gathered} \sqrt{mn}=\sqrt{\left(\tan \left(\theta \right)+\sin \left(\theta \right)\right)\left(\tan \left(\theta \right)-\sin (\theta \right))}\because \left[(a-b)(a+b)=a^2-b^2\right] \\ \Rightarrow \sqrt{mn}=\sqrt{{\tan }^2\left(\theta \right)-{\sin }^2\left(\theta \right)} \\ \Rightarrow \sqrt{mn}=\sqrt{\frac{{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}-{\sin }^2\left(\theta \right)}\because \left[\tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}\right] \\ \Rightarrow \sqrt{mn}=\sqrt{\frac{{\sin }^2\left(\theta \right)\left[1-{\cos }^2\left(\theta \right)\right]}{{\cos }^2\left(\theta \right)}} \\ \Rightarrow \sqrt{mn}=\sqrt{{\tan }^2\left(\theta \right){\sin }^2\left(\theta \right)}\because \left[\tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)},{\sin }^2\left(x\right)+{\cos }^1\left(x\right)=1\right] \\ \Rightarrow mn=\tan \left(\theta \right)\sin \left(\theta \right) \end{gathered}$$

$\Rightarrow m^2-n^2=4\sqrt{mn}$

Hence, option C is correct.