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  4. The value of sec[ tan-1(b+a/b-a)- tan-1(a/b)] =

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Q. The value of $\sec\left[\tan^{-1}\frac{b+a}{b-a}-\tan^{-1}\frac{a}{b}\right]$ =

Inverse Trigonometric Functions

Solution:

$tan^{-1} \frac{b+a}{b-a} - tan^{-1} \frac{a}{b} = tan^{-1}\frac{ \frac{b+a}{b-a}-\frac{a}{b}}{1+ \frac{b+a}{b-a}\cdot\frac{a}{b}} $
$ = tan^{-1} \frac{b^{2}+ab-ab+a^{2}}{b^{2}-ab+ab+a^{2}} $
$= tan^{-1} \frac{a^{2}+b^{2}}{a^{2}+b^{2}} $
$ = tan^{-1} = \frac{\pi}{4} $
$\therefore $ required value $= sec\left( \frac{\pi }{4}\right) = \sqrt{2}$