If $\theta$ is the acute angle between the curves $x^2+y^2=2020 \sqrt{2}$ and $x^2-y^2=2020$, then $\frac{\sin \theta+\cos \theta}{\tan \theta}$ is equal to

• Correct answer
  A

  $\sqrt{2}$
• B

  $\frac{3+\sqrt{3}}{2}$
• C

  $\frac{3+\sqrt{3}}{4}$
• D

  $\frac{3+\sqrt{3}}{6}$

Solution

Adding Eqs. (i) and (ii), we get $$ \begin{aligned} & 2 x^2=2020(\sqrt{2}+1) \\ & \therefore \quad x^2=1010(\sqrt{2}+1), y^2=1010(\sqrt{2}-1) \\ & \therefore \quad x^2 y^2=(1010)^2\left\{(\sqrt{2})^2-(1)^2\right\} \\ & =(1010)^2 \\ & \therefore \quad x y= \pm 1010 \\ & \because \quad x^2+y^2=2020 \sqrt{2}, x^2-y^2=2020 \\ & \Rightarrow \quad 2 x+2 y y^{\prime}=0 \text { and } 2 x-2 y y^{\prime}=0 \\ & \Rightarrow \quad y^{\prime}=\frac{-x}{y}=m_1 \text { (Let) } \\ & \Rightarrow \quad y^{\prime}=\frac{x}{y}=m_2 \text { (let) } \\ & \therefore \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{-\frac{x}{y}-\left(+\frac{x}{y}\right)}{1+\frac{x}{y}\left(\frac{-x}{y}\right)}\right| \\ & =\left|\frac{-2 x}{y} \times \frac{y^2}{y^2-x^2}\right| \\ & =\left|\frac{-2 x y}{y^2-x^2}\right|=\left|\frac{-2( \pm 1010)}{-2020}\right|=| \pm 1|=1 \\ & \end{aligned} $$ $\begin{aligned} \therefore \quad \theta & =\tan ^{-1}(1)=\pi / 4 \\ \therefore \frac{\sin \theta+\cos \theta}{\tan \theta} & =\frac{\sin (\pi / 4)+\cos (\pi / 4)}{\tan (\pi / 4)} \\ & =\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{1}=\frac{2}{\sqrt{2}}=\sqrt{2}\end{aligned}$

Hence, the correct answer is option: 'A'.