Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{a} \sin \theta=\mathrm{b} \cos \theta$, where $\mathrm{a}, \mathrm{b} \neq 0$, then $\mathrm{a} \cos 2 \theta+\mathrm{b} \sin 2 \theta=$
Options:
Solution:
2245 Upvotes
Verified Answer
The correct answer is:
a
$\begin{aligned} & a \sin \theta=b \cos \theta \Rightarrow \tan \theta=\frac{b}{a} \\ & a \cos 2 \theta+b \sin 2 \theta \\ & =a\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)+b\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) \\ & =a\left[\frac{1-\left(\frac{b^2}{a^2}\right)}{1+\left(\frac{b^2}{a^2}\right)}\right]+b\left[\frac{2\left(\frac{b}{a}\right)}{1+\left(\frac{b^2}{a^2}\right)}\right]=a\left(\frac{a^2-b^2}{a^2+b^2}\right)+b\left(\frac{2 b}{a} \times \frac{a^2}{a^2+b^2}\right) \\ & =\frac{a\left(a^2-b^2\right)}{a^2+b^2}+\frac{b(2 a b)}{a^2+b^2}=\frac{a^3-a b^2+2 a b^2}{a^2+b^2} \\ & =\frac{a^3+a b^2}{a^2+b^2}=\frac{a\left(a^2+b^2\right)}{a^2+b^2}=a\end{aligned}$
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.