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NCERT Exemplar · Q5

Q.If tan⁡x=ba\tan x = \dfrac{b}{a}, then find the value of a+ba−b+a−ba+b\sqrt{\dfrac{a + b}{a - b}} + \sqrt{\dfrac{a - b}{a + b}}.

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Rationalise each surd over the common denominator a2−b2\sqrt{a^2-b^2}; the two numerators add directly, giving 2aa2−b2\dfrac{2a}{\sqrt{a^2-b^2}}.

We are given tan⁡x=ba\tan x=\dfrac{b}{a} and must evaluate

E=a+ba−b+a−ba+bE=\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}

Since tan⁡x=ba\tan x=\dfrac{b}{a} is a real, finite value, a+ba+b and a−ba-b both keep the same sign as aa, so the ratios under each root are positive and the surds are well defined.

Step 1 — Rationalise each term.

Multiply the numerator and denominator of each surd's argument by the missing factor so that both denominators become (a+b)(a−b)=a2−b2\sqrt{(a+b)(a-b)}=\sqrt{a^2-b^2}:

a+ba−b=(a+b)(a+b)(a−b)(a+b)=a+ba2−b2\sqrt{\frac{a+b}{a-b}}=\sqrt{\frac{(a+b)(a+b)}{(a-b)(a+b)}}=\frac{a+b}{\sqrt{a^2-b^2}} …

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