If x= b a , find the value of a+b a - b + a - b a+b . - Vidyadip

Question

If $\text{x}=\frac{\text{b}}{\text{a}},$ find the value of $\sqrt{\frac{\text{a+b}}{\text{a - b}}}+\sqrt{\frac{\text{a - b}}{\text{a+b}}}.$

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Answer

Given that: $\tan\text{x}=\frac{\text{b}}{\text{a}}$ $\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}=\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}$ $\frac{\text{a + b + a}-\text{b}}{\sqrt{(\text{a}-\text{b})(\text{a + b})}}=\frac{2\text{a}}{\sqrt{\text{a}^2-\text{b}^2}}=\frac{2\text{a}}{\text{a}\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}}$ $\Big[\because\tan\text{x}=\frac{\text{b}}{\text{a}}\Big]$ $=\frac{2}{1-\tan^2\text{x}}$ $=\frac{2}{\sqrt{1-\frac{\sin^2\text{x}}{\cos^2\text{x}}}}=\frac{2}{\frac{\sqrt{\cos^2\text{x}-\sin^2\text{x}}}{\cos\text{x}}}$ $=\frac{2\cos\text{x}}{\sqrt{\cos2\text{x}}}$ $[\because\cos^2\text{x}-\sin^2\text{x}=\cos2\text{x}]$ Hence, $\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}=\frac{2\cos\text{x}}{\sqrt{\cos2\text{x}}}$

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