Search any question & find its solution
Question:
Answered & Verified by Expert
If the lines given by $\left(x^2+y^2\right) \sin ^2 \alpha=(x \cos \alpha-y \sin \alpha)^2$ are perpendicular to each other, then $\sin ^2 \alpha+\tan ^2 \alpha=$
Options:
Solution:
1928 Upvotes
Verified Answer
The correct answer is:
$\frac{3}{2}$
Given pair of lines
$$
\begin{aligned}
& \left(x^2+y^2\right) \sin ^2 \alpha=(x \cos \alpha-y \sin \alpha)^2 \\
& \Rightarrow x^2 \cos 2 \alpha+y^2(0)-2 x y \sin \alpha \cos \alpha=0
\end{aligned}
$$
Since lines are perpendicular.
$$
\begin{aligned}
& \therefore \text { Coefficient of } x^2+\text { coefficient of } y^2=0 \\
& \Rightarrow \cos 2 \alpha=0
\end{aligned}
$$
$\begin{aligned} & \Rightarrow 1-2 \sin ^2 \alpha=0 \\ & \Rightarrow \sin ^2 \alpha=\frac{1}{2} \\ & \text { and } \cos ^2 \alpha=1-\frac{1}{2}=\frac{1}{2} \\ & \therefore \tan ^2 \alpha=\frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)}=1 \\ & \therefore \sin ^2 \alpha+\tan ^2 \alpha=1+\frac{1}{2}=\frac{3}{2}\end{aligned}$
$$
\begin{aligned}
& \left(x^2+y^2\right) \sin ^2 \alpha=(x \cos \alpha-y \sin \alpha)^2 \\
& \Rightarrow x^2 \cos 2 \alpha+y^2(0)-2 x y \sin \alpha \cos \alpha=0
\end{aligned}
$$
Since lines are perpendicular.
$$
\begin{aligned}
& \therefore \text { Coefficient of } x^2+\text { coefficient of } y^2=0 \\
& \Rightarrow \cos 2 \alpha=0
\end{aligned}
$$
$\begin{aligned} & \Rightarrow 1-2 \sin ^2 \alpha=0 \\ & \Rightarrow \sin ^2 \alpha=\frac{1}{2} \\ & \text { and } \cos ^2 \alpha=1-\frac{1}{2}=\frac{1}{2} \\ & \therefore \tan ^2 \alpha=\frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)}=1 \\ & \therefore \sin ^2 \alpha+\tan ^2 \alpha=1+\frac{1}{2}=\frac{3}{2}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.