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Evaluate: `Cot{Sec^-1(-13/5)}`

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Question

Evaluate:

`cot{sec^-1(-13/5)}`

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Solution

`cot{sec^-1(-13/5)}=cot{sec^-1(pi-13/5)}`

`=-cot{sec^-1(13/5)}`

`=-cot{tan^-1(sqrt(1-(5/13)^3)/(5/13))}`

`=-cot{tan^-1(12/5)}`

`=-cot{cot^-1(5/12)}`

`=-5/12`

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Chapter 3: Inverse Trigonometric Functions - Exercise 4.09 [Page 58]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 3 Inverse Trigonometric Functions
Exercise 4.09 | Q 1.3 | Page 58
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