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AP DSC School Assistant Math Model Test 3

Section: Subject Specific

Question : 101 of 160

Marks: +1, -0

What is

\tan ^{-1}\left(\frac{a}{b}\right)-\tan ^{-1}\left(\frac{a-b}{a+b}\right)

equal to?

Solution:

Concept:

Use the formula for difference of inverse tangents:

\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)

, provided

xy \neq -1

Explanation:

Let

x = \frac{a}{b}

and

y = \frac{a-b}{a+b}

First compute

x - y

\frac{a}{b} - \frac{a-b}{a+b} = \frac{a(a+b) - b(a-b)}{b(a+b)} = \frac{a^2+ab - ab + b^2}{b(a+b)} = \frac{a^2+b^2}{b(a+b)}

Next compute

1 + xy

1 + \frac{a}{b}\cdot\frac{a-b}{a+b} = 1 + \frac{a(a-b)}{b(a+b)} = \frac{b(a+b) + a(a-b)}{b(a+b)} = \frac{ab+b^2 + a^2 - ab}{b(a+b)} = \frac{a^2+b^2}{b(a+b)}

Now apply the formula:

\tan^{-1}\left(\frac{a}{b}\right) - \tan^{-1}\left(\frac{a-b}{a+b}\right) = \tan^{-1}\left(\frac{\frac{a^2+b^2}{b(a+b)}}{\frac{a^2+b^2}{b(a+b)}}\right) = \tan^{-1}(1) = \frac{\pi}{4}

Answer:

The value is

\frac{\pi}{4}

, which corresponds to option B.

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