Search any question & find its solution
Question:
Answered & Verified by Expert
If $a=\cos 2 \alpha+i \sin 2 \alpha, b=\cos 2 \beta+i \sin 2 \beta$, then $\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}=$
Options:
Solution:
1944 Upvotes
Verified Answer
The correct answer is:
$2 \cos (\alpha-\beta)$
We have, $a=\cos 2 \alpha+i \sin 2 \alpha$
and $b=\cos 2 \beta+i \sin 2 \beta$
$\frac{a}{b}=\frac{\operatorname{Cis} 2 \alpha}{\operatorname{Cis} 2 \beta}=\operatorname{Cis}(2 \alpha-2 \beta)$
So, $\sqrt{\frac{a}{b}}=\operatorname{Cis}(\alpha-\beta)=\cos (\alpha-\beta)+i \sin (\alpha-\beta)$
and $\sqrt{\frac{b}{a}}=\cos (\alpha-\beta)-i \sin (\alpha-\beta)$
Now, $\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}=\cos (\alpha-\beta)+i \sin (\alpha-\beta)$
$+\cos (\alpha-\beta)-i \sin (\alpha-\beta)=2 \cos (\alpha-\beta)$
and $b=\cos 2 \beta+i \sin 2 \beta$
$\frac{a}{b}=\frac{\operatorname{Cis} 2 \alpha}{\operatorname{Cis} 2 \beta}=\operatorname{Cis}(2 \alpha-2 \beta)$
So, $\sqrt{\frac{a}{b}}=\operatorname{Cis}(\alpha-\beta)=\cos (\alpha-\beta)+i \sin (\alpha-\beta)$
and $\sqrt{\frac{b}{a}}=\cos (\alpha-\beta)-i \sin (\alpha-\beta)$
Now, $\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}=\cos (\alpha-\beta)+i \sin (\alpha-\beta)$
$+\cos (\alpha-\beta)-i \sin (\alpha-\beta)=2 \cos (\alpha-\beta)$
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.