Test Index

UPSC NDA I 2015 Mathematics Paper

Question Numbers: 40-42

Consider the following statement and answer the questions

Consider,

x=4tan^{-1}(\frac{1}{5}), y= tan^{-1}(\frac{1}{70}) and z= tan^{-1}(\frac{1}{99})

Question : 41 of 120

Marks: +1, -0

What is x – y equal to?

Solution:

Concept:

The formula for difference of inverse tangents:

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

, for

a>0, b>0

Explanation:

From the previous part of the problem, we have

x = \tan^{-1} \left( \frac{120}{119} \right)

and

y = \tan^{-1} \left( \frac{1}{70} \right)

We need

x - y

. Apply the formula with

a = \frac{120}{119}

b = \frac{1}{70}

x - y = \tan^{-1} \left( \frac{ \frac{120}{119} - \frac{1}{70} }{ 1 + \frac{120}{119} \cdot \frac{1}{70} } \right)

Compute numerator:

\frac{120}{119} - \frac{1}{70} = \frac{120 \cdot 70 - 119}{119 \cdot 70} = \frac{8400 - 119}{8330} = \frac{8281}{8330}

Compute denominator:

1 + \frac{120}{119 \cdot 70} = 1 + \frac{120}{8330} = \frac{8330 + 120}{8330} = \frac{8450}{8330}

Thus the fraction simplifies to

\frac{8281/8330}{8450/8330} = \frac{8281}{8450}

Therefore,

x - y = \tan^{-1} \left( \frac{8281}{8450} \right)

Answer:

\tan^{-1} \left( \frac{8281}{8450} \right)

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