show that equation of the line passing through (acos^3 theeta,asin^3theeta) and perpendicular to the line x sec theeta + ycosec theeta = a is x cos theeta -y sin theeta = a cos2 theeta

Dear Student ,Varshini
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Let} \mathrm{co}-\mathrm{ordinates} \mathrm{of} \mathrm{point} \mathrm{P} \mathrm{be} \left(a{\cos }^3\left(\theta \right), a{\sin }^3\left(\theta \right)\right) \mathrm{and} \mathrm{slope} \mathrm{of} \mathrm{required} \mathrm{line} \mathrm{L} \mathrm{be} \mathrm{m} \\ \mathrm{Given} \mathrm{equation} \mathrm{of} \mathrm{line} \mathrm{is} {\mathrm{L}}_1: x\sec \left(\theta \right)+ycsc\left(\theta \right)=a \\ \mathrm{Now} \mathrm{slope} \mathrm{of} \mathrm{any} \mathrm{line} \mathrm{is} \mathrm{given} \mathrm{by} \raisebox{1ex}{$-\left(\mathrm{Coefficient} \mathrm{of} \mathrm{x}\right)$}\!\left/ \!\raisebox{-1ex}{$\left(\mathrm{Coefficient} \mathrm{of} \mathrm{y}\right)$}\right. \\ \Rightarrow \mathrm{Slope} \mathrm{of} {\mathrm{L}}_1=m_1=-\left\{\raisebox{1ex}{$\sec \left(\theta \right)$}\!\left/ \!\raisebox{-1ex}{$\csc \left(\theta \right)$}\right.\right\} \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \sin \left(\theta \right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\csc \left(\theta \right)$}\right. \\ \cos \left(\theta \right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$sec\left(\theta \right)$}\right. \\ \tan \left(\theta \right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\cot \left(\theta \right)$}\right. \\ \tan \left(\theta \right)=\raisebox{1ex}{$\sin \left(\theta \right)$}\!\left/ \!\raisebox{-1ex}{$\cos \left(\theta \right)$}\right. \\ \Rightarrow m_1=-\left\{\raisebox{1ex}{$\sin \left(\theta \right)$}\!\left/ \!\raisebox{-1ex}{$cos\left(\theta \right)$}\right.\right\} \\ \Rightarrow m_1=-\tan \left(\theta \right) \\ \mathrm{Since} {\mathrm{L}}_1 \mathrm{and} \mathrm{L} \mathrm{are} \mathrm{perpendicular} \mathrm{to} \mathrm{each} \mathrm{other} \\ \Rightarrow m=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$m_1$}\right. \\ \Rightarrow m=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$\left\{-\tan \left(\theta \right)\right\}$}\right. \\ \Rightarrow m=\cot \left(\theta \right) \\ \mathrm{Equation} \mathrm{of} \mathrm{any} \mathrm{line} \mathrm{passing} \mathrm{through} \mathrm{point} \left({\mathrm{x}}_1,{\mathrm{y}}_1\right) \mathrm{and} \mathrm{having} \mathrm{slope} \mathrm{m} \mathrm{is} \mathrm{given} \mathrm{by} \\ \left(y-y_1\right)=m\left(x-x_1\right) \\ \mathrm{Hence} \mathrm{equation} \mathrm{of} \mathrm{line} \mathrm{L} \mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by} \\ \left\{y- a{\sin }^3\left(\theta \right)\right\}=\cot \left(\theta \right)\left\{x-a{\cos }^3\left(\theta \right)\right\} \\ \Rightarrow \left\{y- a{\sin }^3\left(\theta \right)\right\}=\left\{\raisebox{1ex}{$\cos \left(\theta \right)$}\!\left/ \!\raisebox{-1ex}{$\sin \left(\theta \right)$}\right.\right\}\left\{x-a{\cos }^3\left(\theta \right)\right\} \\ \Rightarrow \sin \left(\theta \right)\left\{y- a{\sin }^3\left(\theta \right)\right\}=\cos \left(\theta \right)\left\{x-a{\cos }^3\left(\theta \right)\right\} \\ \Rightarrow y\sin \left(\theta \right)- a{\sin }^4\left(\theta \right)=x\cos \left(\theta \right)-a{\cos }^4\left(\theta \right) \\ \Rightarrow x\cos \left(\theta \right)-y\sin \left(\theta \right)=a{\cos }^4\left(\theta \right)- a{\sin }^4\left(\theta \right) \\ \Rightarrow x\cos \left(\theta \right)-y\sin \left(\theta \right)=a\left\{{\cos }^4\left(\theta \right)- {\sin }^4\left(\theta \right)\right\} \\ \Rightarrow x\cos \left(\theta \right)-y\sin \left(\theta \right)=a\left[{\left\{{\cos }^2\left(\theta \right)\right\}}^2- {\left\{{\sin }^2\left(\theta \right)\right\}}^2\right] \\ \Rightarrow x\cos \left(\theta \right)-y\sin \left(\theta \right)=a\left[\left\{{\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right)\right\}\left\{{\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)\right\}\right] \left\{Since a^2-b^2=\left(a-b\right)\left(a+b\right)\right\} \\ \mathrm{We} \mathrm{know} \mathrm{that} {\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)=1 and {\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right)=\cos \left(2\theta \right) \\ \mathbf{\Rightarrow }\mathit{x}\mathit{c}\mathit{o}\mathit{s}\left(\mathbf{\theta }\right)\mathbf{-}\mathit{y}\mathit{s}\mathit{i}\mathit{n}\left(\mathbf{\theta }\right)\mathbf{=}\mathit{a}\mathit{c}\mathit{o}\mathit{s}\left(\mathbf{2}\mathbf{\theta }\right)\mathbf{ }\mathbf{Hence}\mathbf{ }\mathbf{proved} \end{gathered}$$

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards