MHT CET · Maths · Inverse Trigonometric Functions
The value of \(\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right)\) is
- A \(\frac{5}{17}\)
- B \(\frac{6}{17}\)
- C \(\frac{3}{17}\)
- D \(\frac{4}{17}\)
Answer & Solution
Correct Answer
(B) \(\frac{6}{17}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right) \\ & =\cot \left(\sin ^{-1} \frac{3}{5}+\tan ^{-1} \frac{2}{3}\right) \\ & =\cot \left(\tan ^{-1} \frac{\frac{3}{5}}{\sqrt{1-\left(\frac{3}{5}\right)^2}}+\tan ^{-1} \frac{2}{3}\right) \\ & =\cot \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right) \\ & =\cot \left[\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}}\right)\right]\end{aligned}\)
\(\begin{aligned} & =\cot \left[\tan ^{-1}\left(\frac{17}{6}\right)\right] \\ & =\cot \left[\cot ^{-1}\left(\frac{6}{17}\right)\right]=\frac{6}{17}\end{aligned}\)
\(\begin{aligned} & =\cot \left[\tan ^{-1}\left(\frac{17}{6}\right)\right] \\ & =\cot \left[\cot ^{-1}\left(\frac{6}{17}\right)\right]=\frac{6}{17}\end{aligned}\)
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