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$$
\text { If } \tan x=\frac{b}{a} \text {, then find the value of } \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}} \text {. }
$$
\text { If } \tan x=\frac{b}{a} \text {, then find the value of } \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}} \text {. }
$$
Solution:
1442 Upvotes
Verified Answer
We have, $\tan x=\frac{b}{a}$
$$
\begin{aligned}
&\text { Now, } \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\frac{\sqrt{(a+b)^2}+\sqrt{(a-b)^2}}{\sqrt{(a-b)(a+b)}} \\
&=\frac{(a+b)+(a-b)}{\sqrt{a^2-b^2}}=\frac{2 a}{\sqrt{a^2-b^2}}=\frac{2 a}{\sqrt[a]{1-\left(\frac{b}{a}\right)^2}} \\
&=\frac{2}{\sqrt{1-\tan ^2 x}}=\frac{2 \cos x}{\sqrt{\cos ^2 x-\sin ^2 x}}\left[\because \cos 2 x=\cos ^2 x-\sin ^2 x\right] \\
&=\frac{2 \cos x}{\sqrt{\cos 2 x}}
\end{aligned}
$$
$$
\begin{aligned}
&\text { Now, } \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\frac{\sqrt{(a+b)^2}+\sqrt{(a-b)^2}}{\sqrt{(a-b)(a+b)}} \\
&=\frac{(a+b)+(a-b)}{\sqrt{a^2-b^2}}=\frac{2 a}{\sqrt{a^2-b^2}}=\frac{2 a}{\sqrt[a]{1-\left(\frac{b}{a}\right)^2}} \\
&=\frac{2}{\sqrt{1-\tan ^2 x}}=\frac{2 \cos x}{\sqrt{\cos ^2 x-\sin ^2 x}}\left[\because \cos 2 x=\cos ^2 x-\sin ^2 x\right] \\
&=\frac{2 \cos x}{\sqrt{\cos 2 x}}
\end{aligned}
$$
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