Hard A-Level Reciprocal and inverse trig functions Questions

Challenging, exam-style A-Level Reciprocal and inverse trig functions questions with worked solutions. Stretch yourself on the hardest reciprocal trig, trig equations, identities, inverse trig problems.

reciprocal trigtrig equationsidentitiesinverse trigdifferentiationexact values
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
What is the range of arccos⁡x\arccos x?
Show worked solution

Worked solution

  1. Recall the relevant fact or definition

    arccos⁡x∈[0,π]\arccos x\in[0,\pi]

    Bring the key definition, identity or range to mind.

  2. Compare each option against this fact

    test each option in turn\text{test each option in turn}

    Only one option matches exactly.

  3. Recall the reciprocal function definitions

    sec⁡x=1cos⁡x, csc⁡x=1sin⁡x, cot⁡x=1tan⁡x\sec x=\dfrac{1}{\cos x},\ \csc x=\dfrac{1}{\sin x},\ \cot x=\dfrac{1}{\tan x}

    The reciprocal trig functions are built from cosine, sine and tangent.

  4. Recall the Pythagorean identities

    1+tan⁡2x=sec⁡2x,1+cot⁡2x=csc⁡2x1+\tan^2 x=\sec^2 x,\quad 1+\cot^2 x=\csc^2 x

    These come from dividing sin^2 x + cos^2 x = 1 by cos^2 x or by sin^2 x.

  5. Note the ranges of the inverse trig functions

    arcsin⁡x∈[−π2,π2], arccos⁡x∈[0,π], arctan⁡x∈(−π2,π2)\arcsin x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right],\ \arccos x\in[0,\pi],\ \arctan x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)

    Restricting the range makes each trig function one-to-one and invertible.

  6. Relate the angle to the unit circle

    x2+y2=1x^2+y^2=1

    Exact trig ratios come from the unit circle and the special triangles.

  7. Recall the periodicity of the functions

    sec⁡(x+2π)=sec⁡x, cot⁡(x+π)=cot⁡x\sec(x+2\pi)=\sec x,\ \cot(x+\pi)=\cot x

    Periodicity produces further solutions across a full interval.

  8. Confirm any domain restrictions

    cos⁡x≠0, sin⁡x≠0\cos x\neq 0,\ \sin x\neq 0

    Reciprocal functions are undefined where the denominator vanishes.

  9. Cross-check using an equivalent form

    tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}

    Rewriting in an equivalent form provides a useful check.

  10. Recall the graph shape of the function

    y=sec⁡x has vertical asymptotes where cos⁡x=0y=\sec x\ \text{has vertical asymptotes where}\ \cos x=0

    Knowing the graph helps locate every solution.

  11. Evaluate behaviour at key angles

    cos⁡0=1, cos⁡π2=0, sin⁡0=0\cos 0=1,\ \cos\tfrac{\pi}{2}=0,\ \sin 0=0

    Key angles anchor the exact values.

  12. Relate to the corresponding forward trig function

    sin⁡(arcsin⁡x)=x, tan⁡(arctan⁡x)=x\sin(\arcsin x)=x,\ \tan(\arctan x)=x

    Inverse and forward functions undo one another on the correct domain.

  13. Check the boundary behaviour

    −1≤sin⁡x≤1, −1≤cos⁡x≤1-1\le\sin x\le 1,\ -1\le\cos x\le 1

    Endpoints often reveal excluded or limiting values.

  14. Eliminate impossible cases

    reject any value outside the valid range\text{reject any value outside the valid range}

    Removing impossible cases narrows down the answer.

  15. Select the correct option

    0≤arccos⁡x≤π0\le\arccos x\le\pi

    This option matches the recalled fact.

Answer
0≤arccos⁡x≤π0\le\arccos x\le\pi
Question 2
8 markschallenging
Which of the following is equal to cot⁡x\cot x for all valid xx?
Show worked solution

Worked solution

  1. Recall the relevant fact or definition

    cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x}

    Bring the key definition, identity or range to mind.

  2. Compare each option against this fact

    test each option in turn\text{test each option in turn}

    Only one option matches exactly.

  3. Recall the reciprocal function definitions

    sec⁡x=1cos⁡x, csc⁡x=1sin⁡x, cot⁡x=1tan⁡x\sec x=\dfrac{1}{\cos x},\ \csc x=\dfrac{1}{\sin x},\ \cot x=\dfrac{1}{\tan x}

    The reciprocal trig functions are built from cosine, sine and tangent.

  4. Recall the Pythagorean identities

    1+tan⁡2x=sec⁡2x,1+cot⁡2x=csc⁡2x1+\tan^2 x=\sec^2 x,\quad 1+\cot^2 x=\csc^2 x

    These come from dividing sin^2 x + cos^2 x = 1 by cos^2 x or by sin^2 x.

  5. Note the ranges of the inverse trig functions

    arcsin⁡x∈[−π2,π2], arccos⁡x∈[0,π], arctan⁡x∈(−π2,π2)\arcsin x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right],\ \arccos x\in[0,\pi],\ \arctan x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)

    Restricting the range makes each trig function one-to-one and invertible.

  6. Relate the angle to the unit circle

    x2+y2=1x^2+y^2=1

    Exact trig ratios come from the unit circle and the special triangles.

  7. Recall the periodicity of the functions

    sec⁡(x+2π)=sec⁡x, cot⁡(x+π)=cot⁡x\sec(x+2\pi)=\sec x,\ \cot(x+\pi)=\cot x

    Periodicity produces further solutions across a full interval.

  8. Confirm any domain restrictions

    cos⁡x≠0, sin⁡x≠0\cos x\neq 0,\ \sin x\neq 0

    Reciprocal functions are undefined where the denominator vanishes.

  9. Cross-check using an equivalent form

    tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}

    Rewriting in an equivalent form provides a useful check.

  10. Recall the graph shape of the function

    y=sec⁡x has vertical asymptotes where cos⁡x=0y=\sec x\ \text{has vertical asymptotes where}\ \cos x=0

    Knowing the graph helps locate every solution.

  11. Evaluate behaviour at key angles

    cos⁡0=1, cos⁡π2=0, sin⁡0=0\cos 0=1,\ \cos\tfrac{\pi}{2}=0,\ \sin 0=0

    Key angles anchor the exact values.

  12. Relate to the corresponding forward trig function

    sin⁡(arcsin⁡x)=x, tan⁡(arctan⁡x)=x\sin(\arcsin x)=x,\ \tan(\arctan x)=x

    Inverse and forward functions undo one another on the correct domain.

  13. Check the boundary behaviour

    −1≤sin⁡x≤1, −1≤cos⁡x≤1-1\le\sin x\le 1,\ -1\le\cos x\le 1

    Endpoints often reveal excluded or limiting values.

  14. Eliminate impossible cases

    reject any value outside the valid range\text{reject any value outside the valid range}

    Removing impossible cases narrows down the answer.

  15. Select the correct option

    cos⁡xsin⁡x\dfrac{\cos x}{\sin x}

    This option matches the recalled fact.

Answer
cos⁡xsin⁡x\dfrac{\cos x}{\sin x}
Question 3
8 markschallenging
Which function has derivative 11+x2\dfrac{1}{1+x^2}?
Show worked solution

Worked solution

  1. Recall the relevant fact or definition

    ddxarctan⁡x=11+x2\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^2}

    Bring the key definition, identity or range to mind.

  2. Compare each option against this fact

    test each option in turn\text{test each option in turn}

    Only one option matches exactly.

  3. Recall the reciprocal function definitions

    sec⁡x=1cos⁡x, csc⁡x=1sin⁡x, cot⁡x=1tan⁡x\sec x=\dfrac{1}{\cos x},\ \csc x=\dfrac{1}{\sin x},\ \cot x=\dfrac{1}{\tan x}

    The reciprocal trig functions are built from cosine, sine and tangent.

  4. Recall the Pythagorean identities

    1+tan⁡2x=sec⁡2x,1+cot⁡2x=csc⁡2x1+\tan^2 x=\sec^2 x,\quad 1+\cot^2 x=\csc^2 x

    These come from dividing sin^2 x + cos^2 x = 1 by cos^2 x or by sin^2 x.

  5. Note the ranges of the inverse trig functions

    arcsin⁡x∈[−π2,π2], arccos⁡x∈[0,π], arctan⁡x∈(−π2,π2)\arcsin x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right],\ \arccos x\in[0,\pi],\ \arctan x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)

    Restricting the range makes each trig function one-to-one and invertible.

  6. Relate the angle to the unit circle

    x2+y2=1x^2+y^2=1

    Exact trig ratios come from the unit circle and the special triangles.

  7. Recall the periodicity of the functions

    sec⁡(x+2π)=sec⁡x, cot⁡(x+π)=cot⁡x\sec(x+2\pi)=\sec x,\ \cot(x+\pi)=\cot x

    Periodicity produces further solutions across a full interval.

  8. Confirm any domain restrictions

    cos⁡x≠0, sin⁡x≠0\cos x\neq 0,\ \sin x\neq 0

    Reciprocal functions are undefined where the denominator vanishes.

  9. Cross-check using an equivalent form

    tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}

    Rewriting in an equivalent form provides a useful check.

  10. Recall the graph shape of the function

    y=sec⁡x has vertical asymptotes where cos⁡x=0y=\sec x\ \text{has vertical asymptotes where}\ \cos x=0

    Knowing the graph helps locate every solution.

  11. Evaluate behaviour at key angles

    cos⁡0=1, cos⁡π2=0, sin⁡0=0\cos 0=1,\ \cos\tfrac{\pi}{2}=0,\ \sin 0=0

    Key angles anchor the exact values.

  12. Relate to the corresponding forward trig function

    sin⁡(arcsin⁡x)=x, tan⁡(arctan⁡x)=x\sin(\arcsin x)=x,\ \tan(\arctan x)=x

    Inverse and forward functions undo one another on the correct domain.

  13. Check the boundary behaviour

    −1≤sin⁡x≤1, −1≤cos⁡x≤1-1\le\sin x\le 1,\ -1\le\cos x\le 1

    Endpoints often reveal excluded or limiting values.

  14. Eliminate impossible cases

    reject any value outside the valid range\text{reject any value outside the valid range}

    Removing impossible cases narrows down the answer.

  15. Select the correct option

    arctan⁡x\arctan x

    This option matches the recalled fact.

Answer
arctan⁡x\arctan x
Question 4
8 markschallenging
What is the domain of arccos⁡x\arccos x?
Show worked solution

Worked solution

  1. Recall the relevant fact or definition

    arccos⁡x requires −1≤x≤1\arccos x\ \text{requires}\ -1\le x\le 1

    Bring the key definition, identity or range to mind.

  2. Compare each option against this fact

    test each option in turn\text{test each option in turn}

    Only one option matches exactly.

  3. Recall the reciprocal function definitions

    sec⁡x=1cos⁡x, csc⁡x=1sin⁡x, cot⁡x=1tan⁡x\sec x=\dfrac{1}{\cos x},\ \csc x=\dfrac{1}{\sin x},\ \cot x=\dfrac{1}{\tan x}

    The reciprocal trig functions are built from cosine, sine and tangent.

  4. Recall the Pythagorean identities

    1+tan⁡2x=sec⁡2x,1+cot⁡2x=csc⁡2x1+\tan^2 x=\sec^2 x,\quad 1+\cot^2 x=\csc^2 x

    These come from dividing sin^2 x + cos^2 x = 1 by cos^2 x or by sin^2 x.

  5. Note the ranges of the inverse trig functions

    arcsin⁡x∈[−π2,π2], arccos⁡x∈[0,π], arctan⁡x∈(−π2,π2)\arcsin x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right],\ \arccos x\in[0,\pi],\ \arctan x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)

    Restricting the range makes each trig function one-to-one and invertible.

  6. Relate the angle to the unit circle

    x2+y2=1x^2+y^2=1

    Exact trig ratios come from the unit circle and the special triangles.

  7. Recall the periodicity of the functions

    sec⁡(x+2π)=sec⁡x, cot⁡(x+π)=cot⁡x\sec(x+2\pi)=\sec x,\ \cot(x+\pi)=\cot x

    Periodicity produces further solutions across a full interval.

  8. Confirm any domain restrictions

    cos⁡x≠0, sin⁡x≠0\cos x\neq 0,\ \sin x\neq 0

    Reciprocal functions are undefined where the denominator vanishes.

  9. Cross-check using an equivalent form

    tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}

    Rewriting in an equivalent form provides a useful check.

  10. Recall the graph shape of the function

    y=sec⁡x has vertical asymptotes where cos⁡x=0y=\sec x\ \text{has vertical asymptotes where}\ \cos x=0

    Knowing the graph helps locate every solution.

  11. Evaluate behaviour at key angles

    cos⁡0=1, cos⁡π2=0, sin⁡0=0\cos 0=1,\ \cos\tfrac{\pi}{2}=0,\ \sin 0=0

    Key angles anchor the exact values.

  12. Relate to the corresponding forward trig function

    sin⁡(arcsin⁡x)=x, tan⁡(arctan⁡x)=x\sin(\arcsin x)=x,\ \tan(\arctan x)=x

    Inverse and forward functions undo one another on the correct domain.

  13. Check the boundary behaviour

    −1≤sin⁡x≤1, −1≤cos⁡x≤1-1\le\sin x\le 1,\ -1\le\cos x\le 1

    Endpoints often reveal excluded or limiting values.

  14. Eliminate impossible cases

    reject any value outside the valid range\text{reject any value outside the valid range}

    Removing impossible cases narrows down the answer.

  15. Select the correct option

    [−1, 1][-1,\ 1]

    This option matches the recalled fact.

Answer
[−1, 1][-1,\ 1]
Question 5
8 markschallenging
The expression csc⁡2x−cot⁡2x\csc^2 x-\cot^2 x simplifies to which of the following?
Show worked solution

Worked solution

  1. Recall the relevant fact or definition

    1+cot⁡2x=csc⁡2x ⇒ csc⁡2x−cot⁡2x=11+\cot^2 x=\csc^2 x\ \Rightarrow\ \csc^2 x-\cot^2 x=1

    Bring the key definition, identity or range to mind.

  2. Compare each option against this fact

    test each option in turn\text{test each option in turn}

    Only one option matches exactly.

  3. Recall the reciprocal function definitions

    sec⁡x=1cos⁡x, csc⁡x=1sin⁡x, cot⁡x=1tan⁡x\sec x=\dfrac{1}{\cos x},\ \csc x=\dfrac{1}{\sin x},\ \cot x=\dfrac{1}{\tan x}

    The reciprocal trig functions are built from cosine, sine and tangent.

  4. Recall the Pythagorean identities

    1+tan⁡2x=sec⁡2x,1+cot⁡2x=csc⁡2x1+\tan^2 x=\sec^2 x,\quad 1+\cot^2 x=\csc^2 x

    These come from dividing sin^2 x + cos^2 x = 1 by cos^2 x or by sin^2 x.

  5. Note the ranges of the inverse trig functions

    arcsin⁡x∈[−π2,π2], arccos⁡x∈[0,π], arctan⁡x∈(−π2,π2)\arcsin x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right],\ \arccos x\in[0,\pi],\ \arctan x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)

    Restricting the range makes each trig function one-to-one and invertible.

  6. Relate the angle to the unit circle

    x2+y2=1x^2+y^2=1

    Exact trig ratios come from the unit circle and the special triangles.

  7. Recall the periodicity of the functions

    sec⁡(x+2π)=sec⁡x, cot⁡(x+π)=cot⁡x\sec(x+2\pi)=\sec x,\ \cot(x+\pi)=\cot x

    Periodicity produces further solutions across a full interval.

  8. Confirm any domain restrictions

    cos⁡x≠0, sin⁡x≠0\cos x\neq 0,\ \sin x\neq 0

    Reciprocal functions are undefined where the denominator vanishes.

  9. Cross-check using an equivalent form

    tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}

    Rewriting in an equivalent form provides a useful check.

  10. Recall the graph shape of the function

    y=sec⁡x has vertical asymptotes where cos⁡x=0y=\sec x\ \text{has vertical asymptotes where}\ \cos x=0

    Knowing the graph helps locate every solution.

  11. Evaluate behaviour at key angles

    cos⁡0=1, cos⁡π2=0, sin⁡0=0\cos 0=1,\ \cos\tfrac{\pi}{2}=0,\ \sin 0=0

    Key angles anchor the exact values.

  12. Relate to the corresponding forward trig function

    sin⁡(arcsin⁡x)=x, tan⁡(arctan⁡x)=x\sin(\arcsin x)=x,\ \tan(\arctan x)=x

    Inverse and forward functions undo one another on the correct domain.

  13. Check the boundary behaviour

    −1≤sin⁡x≤1, −1≤cos⁡x≤1-1\le\sin x\le 1,\ -1\le\cos x\le 1

    Endpoints often reveal excluded or limiting values.

  14. Eliminate impossible cases

    reject any value outside the valid range\text{reject any value outside the valid range}

    Removing impossible cases narrows down the answer.

  15. Select the correct option

    11

    This option matches the recalled fact.

Answer
11

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