alleryStart learning →Value of arccos(-1):
- -π
- π/2
- 0
- π
Show answer and solution
Correct answer: D. π
cos(π) = -1, so arccos(-1) = π.
(Principal range [0, π].)
Indian Navy 10+2 B.Tech Cadet Inverse Trigonometric Functions — practice questions
Explore 28 free Inverse Trigonometric Functions questions in Indian Navy 10+2 B.Tech Cadet Mathematics. Open each question or exercise to read its content, reference answers and available explanation.
Practise Inverse Trigonometric Functions with Quill →Inverse Trigonometric Functions questions and exercises with answers
Showing questions 1–28 of 28.
Value of arccos(-1):Read the answer and worked solution →$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = $Read the answer and worked solution →Sort each inverse function into its domainRead the answer and worked solution →Tap every function whose domain is [−1, 1]Read the answer and worked solution →Match each inverse function to its domainRead the answer and worked solution →Match each inverse function to its principal value branchRead the answer and worked solution →Flip and pair: function with its principal rangeRead the answer and worked solution →Which principal ranges sit around zero, and which run 0 to π?Read the answer and worked solution →Tap every function whose principal range has a point removedRead the answer and worked solution →Name the principal value branchRead the answer and worked solution →Domain checkRead the answer and worked solution →Put the reasoning in order: why an inverse needs a restricted domainRead the answer and worked solution →True or false: existence of inversesRead the answer and worked solution →True or false: reading the tableRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Complete the summary tableRead the answer and worked solution →Complete the rule about restricting a domainRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Principal values worked in the chapterRead the answer and worked solution →Order the steps for finding a principal valueRead the answer and worked solution →True or false: principal valuesRead the answer and worked solution →Match each identity to the condition it needsRead the answer and worked solution →Assertion and reason: identitiesRead the answer and worked solution →Match each mathematician to the year your syllabus givesRead the answer and worked solution →Trace how trigonometry travelledRead the answer and worked solution →True or false: the historical noteRead the answer and worked solution →
Worked practice questions
$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = $
- $\tfrac{\pi}{2}$
- $\tfrac{5\pi}{4}$
- $\tfrac{3\pi}{4}$
- $\pi$
Show answer and solution
Correct answer: D. $\pi$
For $\tan^{-1}(2) + \tan^{-1}(3)$, the product $2\cdot 3 = 6 > 1$
so use $\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\tfrac{x+y}{1-xy}$ (positive case).
This gives $\pi + \tan^{-1}\tfrac{5}{-5} = \pi + \tan^{-1}(-1) = \pi - \tfrac{\pi}{4} = \tfrac{3\pi}{4}$.
Adding $\tan^{-1}(1) = \tfrac{\pi}{4}$: total $= \pi$.
Sort each inverse function into its domain
Read exercise and reference answers
- Items
- Item
- sin⁻¹ x
- Category
- [−1, 1]
- Item
- cos⁻¹ x
- Category
- [−1, 1]
- Item
- cosec⁻¹ x
- Category
- R − (−1, 1)
- Item
- sec⁻¹ x
- Category
- R − (−1, 1)
- Item
- tan⁻¹ x
- Category
- R
- Item
- cot⁻¹ x
- Category
- R
- Categories
- Item
- [−1, 1]
- Item
- R − (−1, 1)
- Item
- R
Only sin⁻¹ and cos⁻¹ accept every value from −1 to 1. cosec⁻¹ and sec⁻¹ need |x| ≥ 1, so the open middle is cut out.
tan⁻¹ and cot⁻¹ accept any real number.
Source: syllabus Maths Part I, Ch 2 Inverse Trigonometric Functions.
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