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Indian Navy Pilot Inverse Trigonometric Functions — practice questions

Explore 28 free Inverse Trigonometric Functions questions in Indian Navy Pilot Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Inverse Trigonometric Functions questions and exercises with answers

Showing questions 1–28 of 28.

If $\sin^{-1}(x) + \sin^{-1}(y) = \dfrac{\pi}{2}$, then $x^2 + y^2$ equals: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

If $\sin^{-1}(x) + \sin^{-1}(y) = \dfrac{\pi}{2}$, then $x^2 + y^2$ equals:

  1. $1$
  2. $\dfrac{1}{2}$
  3. $0$
  4. $2$
Show answer and solution

Correct answer: A. $1$

Let $\sin^{-1}(x) = \alpha$ and $\sin^{-1}(y) = \beta$. So $\alpha + \beta = \dfrac{\pi}{2} \Rightarrow \beta = \dfrac{\pi}{2} - \alpha$. Then $y = \sin\beta = \sin\left(\dfrac{\pi}{2} - \alpha\right) = \cos\alpha$. Since $\sin^2\alpha + \cos^2\alpha = 1$: $x^2 + y^2 = \sin^2\alpha + \cos^2\alpha = 1$.

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$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = $

  1. $\tfrac{\pi}{2}$
  2. $\tfrac{5\pi}{4}$
  3. $\tfrac{3\pi}{4}$
  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 $= \pi$.

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Sort each inverse function into its domain

Read exercise and reference answers
Items
  1. Item
    sin⁻¹ x
    Category
    [−1, 1]
  2. Item
    cos⁻¹ x
    Category
    [−1, 1]
  3. Item
    cosec⁻¹ x
    Category
    R − (−1, 1)
  4. Item
    sec⁻¹ x
    Category
    R − (−1, 1)
  5. Item
    tan⁻¹ x
    Category
    R
  6. Item
    cot⁻¹ x
    Category
    R
Categories
  1. Item
    [−1, 1]
  2. Item
    R − (−1, 1)
  3. 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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