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Inverse Trigonometric Functions · Question 102 of 120

Question 102

MediumOriginal question

What is the value of tan⁡−1(−5)+cot⁡−1(−5)\tan^{-1}(-5) + \cot^{-1}(-5)?

Options for question 102

Show the answer and explanation
CorrectNot quiteAnswer C

Correct answer: C

Explanation

  1. Let θ=tan⁡−1(−5)\theta = \tan^{-1}(-5), in (−π2,0)\left(-\frac{\pi}{2}, 0\right). Then π2−θ\frac{\pi}{2} - \theta lies in (0,π)(0, \pi) and cot⁡(π2−θ)=tan⁡θ=−5\cot\left(\frac{\pi}{2} - \theta\right) = \tan \theta = -5.

  2. So cot⁡−1(−5)=π2−θ\cot^{-1}(-5) = \frac{\pi}{2} - \theta, and the sum is π2\frac{\pi}{2}, as it is for every real input.

Worth noting

−π2-\frac{\pi}{2} writes cot⁡−1(−5)\cot^{-1}(-5) as tan⁡−1(−15)\tan^{-1}\left(-\frac{1}{5}\right), a negative angle off the branch (0,π)(0, \pi).

Concept
Complementary principal values
Typical time
90 s
Your time
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The rule this testsValues of expressions
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Ten questions from Inverse Trigonometric Functions, at your own pace.

Practise this chapter →

Ten questions from Inverse Trigonometric Functions, at your own pace.