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 : 42 of 120

Marks: +1, -0

What is x – y + z equal to?

Solution:

Concept:

Use the formula for sum of two inverse tangents:

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

when

ab<1

Explanation:

From the previous calculation, we have

x - y = \tan^{-1}\left(\frac{49}{50}\right)

Given

z = \tan^{-1}\left(\frac{1}{99}\right)

x - y + z = \tan^{-1}\left(\frac{49}{50}\right) + \tan^{-1}\left(\frac{1}{99}\right)

Apply the sum formula:

\tan^{-1}\left(\frac{49}{50}\right) + \tan^{-1}\left(\frac{1}{99}\right) = \tan^{-1}\left(\frac{\frac{49}{50} + \frac{1}{99}}{1 - \frac{49}{50} \cdot \frac{1}{99}}\right)

Simplify numerator:

\frac{49}{50} + \frac{1}{99} = \frac{49 \cdot 99 + 50}{50 \cdot 99} = \frac{4851 + 50}{4950} = \frac{4901}{4950}

Denominator:

1 - \frac{49}{50} \cdot \frac{1}{99} = 1 - \frac{49}{4950} = \frac{4950 - 49}{4950} = \frac{4901}{4950}

Thus the fraction equals

1

, so

x - y + z = \tan^{-1}(1) = \frac{\pi}{4}

Answer:

\frac{\pi}{4}

(Option D).

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