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Mathematics Question on Properties of Inverse Trigonometric Functions

The value of sec⁡[tan⁡−1b+ab−a−tan⁡−1ab]\sec\left[\tan^{-1}\frac{b+a}{b-a}-\tan^{-1}\frac{a}{b}\right] =

A

2

B

2\sqrt2

C

4

D

1

Answer

2\sqrt2

Explanation

Solution

tan−1b+ab−a−tan−1ab=tan−1b+ab−a−ab1+b+ab−a⋅abtan^{-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−1b2+ab−ab+a2b2−ab+ab+a2= tan^{-1} \frac{b^{2}+ab-ab+a^{2}}{b^{2}-ab+ab+a^{2}} =tan−1a2+b2a2+b2= tan^{-1} \frac{a^{2}+b^{2}}{a^{2}+b^{2}} =tan−1=π4= tan^{-1} = \frac{\pi}{4} ∴\therefore required value =sec(π4)=2= sec\left( \frac{\pi }{4}\right) = \sqrt{2}