Question

Find the value of tan^-1⁡(\(\frac{1}{3}\))+tan^-1⁡(\(\frac{1}{5}\))+tan^-1⁡(\frac{1}{7})[/latex]

(a) tan^-1⁡\((\frac{4}{7})\)

(b) tan^-1⁡\((\frac{9}{7})\)

(c) tan^-1⁡\((\frac{7}{9})\)

(d) tan^-1⁡1

The question was asked during an internship interview.

I would like to ask this question from Properties of Inverse Trigonometric Functions in portion Inverse Trigonometric Functions of Mathematics – Class 12

Answer 1

Right option is (c) tan^-1⁡\((\frac{7}{9})\)

The best I can explain: Using the formula tan^-1⁡x+tan^-1⁡y=tan^-1⁡\(\frac{x+y}{1-xy}\), we get

tan^-1⁡(\(\frac{1}{3}\))+tan^-1⁡(\(\frac{1}{5}\))=tan^-1⁡\(\bigg(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3}×\frac{1}{5}}\bigg)\)

= \(tan^{-1}\bigg(\frac{\frac{8}{15}}{\frac{14}{15}}\bigg)=tan^{-1}⁡(\frac{8}{15}×\frac{15}{14})=tan^{-1}⁡(\frac{4}{7})\)

=\(tan^{-1}(\frac{1}{3})+tan^{-1}(\frac{1}{5})+tan^{-1}⁡(\frac{1}{7})=tan^{-1}(\frac{4}{7})+tan^{-1}⁡(\frac{1}{7})\)

=\(tan^{-1}⁡\bigg(\frac{\frac{4}{7} + \frac{1}{7}}{1-\frac{4}{7}×\frac{1}{7}}\bigg) = tan^{-1}⁡\bigg(\frac{\frac{5}{7}}{\frac{45}{49}}\bigg)=tan^{-1}⁡(\frac{5}{7}×\frac{49}{45})\)

=tan^-1⁡\((\frac{7}{9})\).