Question

\( \tan^{-1}\frac{3}{5} + \tan^{-1}\frac{6}{41} + \tan^{-1}\frac{9}{191} = \)

• A. \( \tan^{-1}\frac{9}{10} \)
• B. \( \tan^{-1}\frac{18}{19} \)
• C. \( \tan^{-1}\frac{3}{191} \)
• D. \( \tan^{-1}\frac{6}{205} \)

Hint:

Look for factors like 17, 13, 19 when simplifying large fractions in inverse trigonometry problems. Simplification at intermediate steps reduces calculation errors.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

We evaluate the sum of inverse tangent functions by grouping terms and using the addition formula step by step.

Step 2: Key Formula or Approach:

tan-1x + tan-1y = tan-1((x + y) / (1 - xy))
for xy < 1.

Step 3: Detailed Explanation:

First Addition:

tan-1(3/5) + tan-1(6/41)

= tan-1(((3/5) + (6/41)) / (1 - (3/5)(6/41)))

= tan-1(((123 + 30)/205) / ((205 - 18)/205))

= tan-1(153/187)

Now simplify:
153 = 9 × 17
187 = 11 × 17

So,
tan-1(153/187) = tan-1(9/11)

Second Addition:

tan-1(9/11) + tan-1(9/191)

= tan-1(((9/11) + (9/191)) / (1 - (9/11)(9/191)))

Numerator:
(9 × 191 + 9 × 11) / (11 × 191)
= 9(191 + 11) / 2101
= 9 × 202 / 2101

Denominator:
(11 × 191 - 81) / 2101
= (2101 - 81) / 2101
= 2020 / 2101

Therefore,
= tan-1((9 × 202 / 2101) / (2020 / 2101))

= tan-1((9 × 202) / 2020)

Since 2020 = 10 × 202,
= tan-1(9/10)

Step 4: Final Answer:

The sum is tan-1(9/10).