$y=\sqrt{\cos 2 x}$ হলে, $\left(y y_{1}\right)^{2}= ?$

Solve: এখানে, $y=\sqrt{\cos 2 x} \Rightarrow y^{2}=\cos 2 x$ উভয় পক্ষকে $x$ -এর সাপেক্ষ অন্তরীকরণ করে পাই,

$\begin{aligned} & 2 y y_{1}=-\sin 2 x .2 \Rightarrow y y_{1}=-\sin 2 x \\ \Rightarrow & \left(y y_{1}\right)^{2}=\sin ^{2} 2 x \quad[\text { squaring both sides } \\ \Rightarrow & \left(y y_{1}\right)^{2}=1-\cos ^{2} 2 x \\ & =1-\left(y^{2}\right)^{2} \quad\left[\because y^{2}=\cos 2 x\right] \\ \therefore \quad & \left(y y_{1}\right)^{2}=1-y^{4} \quad \text { } \end{aligned}$