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Complementary Angles — ICSE Class 9 Maths Important Questions

13 hand-picked ICSE Class 9 Maths important questions for Complementary Angles, each with a full model answer — the formats and topics most likely to appear in your board exam.

13
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6
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32
Total marks
₹0
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Quick answer

In ICSE Class 9, complementary angles give the ratios sin⁡(90∘−A)=cos⁡A\sin(90^\circ-A)=\cos A(90^-A)= A, cos⁡(90∘−A)=sin⁡A\cos(90^\circ-A)=\sin A(90^-A)= A, tan⁡(90∘−A)=cot⁡A\tan(90^\circ-A)=\cot A(90^-A)= A and their reciprocals. High-yield questions evaluate expressions like sin⁡35∘cos⁡55∘\dfrac{\sin35^\circ}{\cos55^\circ}35^/55^, prove identities, and solve sec⁡4A=csc⁡(A−20∘)\sec4A=\csc(A-20^\circ)4A=(A-20^) type equations.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Complementary Angles

In the ICSE Class 9 Maths chapter Trigonometric Ratios of Complementary Angles you use the fact that two angles are complementary when they add to 90∘90^\circ90^. This gives sin⁡(90∘−A)=cos⁡A\sin(90^\circ-A)=\cos A(90^-A)= A, cos⁡(90∘−A)=sin⁡A\cos(90^\circ-A)=\sin A(90^-A)= A, tan⁡(90∘−A)=cot⁡A\tan(90^\circ-A)=\cot A(90^-A)= A, cot⁡(90∘−A)=tan⁡A\cot(90^\circ-A)=\tan A(90^-A)= A, sec⁡(90∘−A)=csc⁡A\sec(90^\circ-A)=\csc A(90^-A)= A and csc⁡(90∘−A)=sec⁡A\csc(90^\circ-A)=\sec A(90^-A)= A, which you apply to evaluate ratios, prove identities and find unknown angles.

Meaning of complementary anglesRatios of complementary angles ($90^\circ-A$)Evaluating trigonometric expressionsProving trigonometric identitiesFinding unknown angles from equations

Key concepts & formulas

Complementary ratios

sin⁡(90∘−A)=cos⁡A\sin(90^\circ-A)=\cos A(90^-A)= A, cos⁡(90∘−A)=sin⁡A\cos(90^\circ-A)=\sin A(90^-A)= A, tan⁡(90∘−A)=cot⁡A\tan(90^\circ-A)=\cot A(90^-A)= A, cot⁡(90∘−A)=tan⁡A\cot(90^\circ-A)=\tan A(90^-A)= A, sec⁡(90∘−A)=csc⁡A\sec(90^\circ-A)=\csc A(90^-A)= A, csc⁡(90∘−A)=sec⁡A\csc(90^\circ-A)=\sec A(90^-A)= A.

Key ratio

sin⁡(90∘−A)cos⁡A=1\dfrac{\sin(90^\circ-A)}{\cos A}=1(90^-A)/ A=1 and cos⁡(90∘−A)sin⁡A=1\dfrac{\cos(90^\circ-A)}{\sin A}=1(90^-A)/ A=1, so any ratio of an angle and its complement's co-ratio equals 111.

Solving for the angle

If sin⁡A=cos⁡B\sin A=\cos BA= B then A+B=90∘A+B=90^\circA+B=90^; likewise tan⁡A=cot⁡B⇒A+B=90∘\tan A=\cot B\Rightarrow A+B=90^\circA= B A+B=90^ and sec⁡A=csc⁡B⇒A+B=90∘\sec A=\csc B\Rightarrow A+B=90^\circA= B A+B=90^ (for acute angles).

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The value of sin⁡35∘cos⁡55∘\dfrac{\sin35^\circ}{\cos55^\circ}35^/55^ is:

  1. (a)

    000

  2. (b)

    111

  3. (c)

    12\dfrac{1}{2}1/2

  4. (d)

    3\sqrt33

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Answer: (b) 111.

Since cos⁡55∘=cos⁡(90∘−35∘)=sin⁡35∘\cos55^\circ=\cos(90^\circ-35^\circ)=\sin35^\circ55^=(90^-35^)=35^, we get sin⁡35∘sin⁡35∘=1\dfrac{\sin35^\circ}{\sin35^\circ}=135^/35^=1.

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Q2MCQEasy1 mark

cos⁡(90∘−A)\cos(90^\circ-A)(90^-A) is equal to:

  1. (a)

    sin⁡A\sin AA

  2. (b)

    cos⁡A\cos AA

  3. (c)

    tan⁡A\tan AA

  4. (d)

    sec⁡A\sec AA

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Answer: (a) sin⁡A\sin AA.

By the complementary-angle rule, cos⁡(90∘−A)=sin⁡A\cos(90^\circ-A)=\sin A(90^-A)= A.

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Q3MCQModerate1 mark

If sec⁡4A=csc⁡(A−20∘)\sec4A=\csc(A-20^\circ)4A=(A-20^), where 4A4A4A is an acute angle, then AAA is:

  1. (a)

    18∘18^\circ18^

  2. (b)

    22∘22^\circ22^

  3. (c)

    24∘24^\circ24^

  4. (d)

    30∘30^\circ30^

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Answer: (b) 22∘22^\circ22^.

sec⁡4A=csc⁡(90∘−4A)\sec4A=\csc(90^\circ-4A)4A=(90^-4A), so csc⁡(90∘−4A)=csc⁡(A−20∘)\csc(90^\circ-4A)=\csc(A-20^\circ)(90^-4A)=(A-20^). Then 90∘−4A=A−20∘⇒5A=110∘⇒A=22∘90^\circ-4A=A-20^\circ\Rightarrow 5A=110^\circ\Rightarrow A=22^\circ90^-4A=A-20^ 5A=110^ A=22^.

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Q4MCQHOTS1 mark

The value of tan⁡48∘ tan⁡23∘ tan⁡42∘ tan⁡67∘\tan48^\circ\,\tan23^\circ\,\tan42^\circ\,\tan67^\circ48^\,23^\,42^\,67^ is:

  1. (a)

    000

  2. (b)

    12\dfrac{1}{2}1/2

  3. (c)

    111

  4. (d)

    222

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Answer: (c) 111.

Pair complements: tan⁡48∘tan⁡42∘=tan⁡48∘cot⁡48∘=1\tan48^\circ\tan42^\circ=\tan48^\circ\cot48^\circ=148^42^=48^48^=1 and tan⁡23∘tan⁡67∘=tan⁡23∘cot⁡23∘=1\tan23^\circ\tan67^\circ=\tan23^\circ\cot23^\circ=123^67^=23^23^=1. Product =1×1=1=1\times1=1=1×1=1.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): tan⁡20∘cot⁡70∘=1\dfrac{\tan20^\circ}{\cot70^\circ}=120^/70^=1.

Reason (R): cot⁡(90∘−θ)=tan⁡θ\cot(90^\circ-\theta)=\tan\theta(90^-)=.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (a) cot⁡70∘=cot⁡(90∘−20∘)=tan⁡20∘\cot70^\circ=\cot(90^\circ-20^\circ)=\tan20^\circ70^=(90^-20^)=20^, so tan⁡20∘tan⁡20∘=1\dfrac{\tan20^\circ}{\tan20^\circ}=120^/20^=1. R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Evaluate: cos⁡40∘sin⁡50∘+sin⁡32∘cos⁡58∘\dfrac{\cos40^\circ}{\sin50^\circ}+\dfrac{\sin32^\circ}{\cos58^\circ}40^/50^+32^/58^.

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sin⁡50∘=sin⁡(90∘−40∘)=cos⁡40∘\sin50^\circ=\sin(90^\circ-40^\circ)=\cos40^\circ50^=(90^-40^)=40^, so the first term =cos⁡40∘cos⁡40∘=1=\dfrac{\cos40^\circ}{\cos40^\circ}=1=40^/40^=1.

cos⁡58∘=cos⁡(90∘−32∘)=sin⁡32∘\cos58^\circ=\cos(90^\circ-32^\circ)=\sin32^\circ58^=(90^-32^)=32^, so the second term =sin⁡32∘sin⁡32∘=1=\dfrac{\sin32^\circ}{\sin32^\circ}=1=32^/32^=1.

∴\therefore value =1+1=2=1+1=2=1+1=2.

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Q7Very ShortModerate2 marks

Without using tables, evaluate sec⁡70∘sin⁡20∘+cos⁡20∘csc⁡70∘\sec70^\circ\sin20^\circ+\cos20^\circ\csc70^\circ70^20^+20^70^.

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sec⁡70∘=sec⁡(90∘−20∘)=csc⁡20∘=1sin⁡20∘\sec70^\circ=\sec(90^\circ-20^\circ)=\csc20^\circ=\dfrac{1}{\sin20^\circ}70^=(90^-20^)=20^=1/20^, so sec⁡70∘sin⁡20∘=1\sec70^\circ\sin20^\circ=170^20^=1.

csc⁡70∘=csc⁡(90∘−20∘)=sec⁡20∘=1cos⁡20∘\csc70^\circ=\csc(90^\circ-20^\circ)=\sec20^\circ=\dfrac{1}{\cos20^\circ}70^=(90^-20^)=20^=1/20^, so cos⁡20∘csc⁡70∘=1\cos20^\circ\csc70^\circ=120^70^=1.

∴\therefore value =1+1=2=1+1=2=1+1=2.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Express cos⁡75∘+csc⁡80∘+tan⁡55∘\cos75^\circ+\csc80^\circ+\tan55^\circ75^+80^+55^ in terms of trigonometric ratios of angles between 0∘0^\circ0^ and 45∘45^\circ45^.

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Use 90∘−θ90^\circ-\theta90^- to bring each angle below 45∘45^\circ45^.

cos⁡75∘=cos⁡(90∘−15∘)=sin⁡15∘\cos75^\circ=\cos(90^\circ-15^\circ)=\sin15^\circ75^=(90^-15^)=15^.

csc⁡80∘=csc⁡(90∘−10∘)=sec⁡10∘\csc80^\circ=\csc(90^\circ-10^\circ)=\sec10^\circ80^=(90^-10^)=10^.

tan⁡55∘=tan⁡(90∘−35∘)=cot⁡35∘\tan55^\circ=\tan(90^\circ-35^\circ)=\cot35^\circ55^=(90^-35^)=35^.

∴cos⁡75∘+csc⁡80∘+tan⁡55∘=sin⁡15∘+sec⁡10∘+cot⁡35∘.\therefore \cos75^\circ+\csc80^\circ+\tan55^\circ=\sin15^\circ+\sec10^\circ+\cot35^\circ.75^+80^+55^=15^+10^+35^.

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Q9Short AnswerModerate3 marks

If tan⁡2A=cot⁡(A−18∘)\tan2A=\cot(A-18^\circ)2A=(A-18^), where 2A2A2A is an acute angle, find the value of AAA.

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cot⁡(A−18∘)=tan⁡(90∘−(A−18∘))=tan⁡(108∘−A)\cot(A-18^\circ)=\tan\big(90^\circ-(A-18^\circ)\big)=\tan(108^\circ-A)(A-18^)=(90^-(A-18^))=(108^-A).

So tan⁡2A=tan⁡(108∘−A)\tan2A=\tan(108^\circ-A)2A=(108^-A).

⇒2A=108∘−A\Rightarrow 2A=108^\circ-A2A=108^-A

⇒3A=108∘\Rightarrow 3A=108^\circ3A=108^

⇒A=36∘.\Rightarrow A=36^\circ.A=36^.

Check: 2A=72∘2A=72^\circ2A=72^ is acute, so the solution is valid.

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Q10Short AnswerHOTS3 marks

Prove that sin⁡(90∘−θ)cos⁡(90∘−θ)tan⁡(90∘−θ)=sin⁡2θ\dfrac{\sin(90^\circ-\theta)\cos(90^\circ-\theta)}{\tan(90^\circ-\theta)}=\sin^2\theta(90^-)(90^-)/(90^-)=^2.

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Convert each complementary ratio.

sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ-\theta)=\cos\theta(90^-)=, cos⁡(90∘−θ)=sin⁡θ\cos(90^\circ-\theta)=\sin\theta(90^-)=, tan⁡(90∘−θ)=cot⁡θ=cos⁡θsin⁡θ\tan(90^\circ-\theta)=\cot\theta=\dfrac{\cos\theta}{\sin\theta}(90^-)==/.

LHS =cos⁡θ⋅sin⁡θcos⁡θsin⁡θ=cos⁡θsin⁡θ×sin⁡θcos⁡θ=sin⁡2θ==\dfrac{\cos\theta\cdot\sin\theta}{\dfrac{\cos\theta}{\sin\theta}}=\cos\theta\sin\theta\times\dfrac{\sin\theta}{\cos\theta}=\sin^2\theta==·/=×/=^2= RHS.

Hence proved.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

Evaluate: sin⁡18∘cos⁡72∘+3(tan⁡10∘tan⁡30∘tan⁡80∘)\dfrac{\sin18^\circ}{\cos72^\circ}+\sqrt3\left(\tan10^\circ\tan30^\circ\tan80^\circ\right)18^/72^+3(10^30^80^).

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First term. cos⁡72∘=cos⁡(90∘−18∘)=sin⁡18∘\cos72^\circ=\cos(90^\circ-18^\circ)=\sin18^\circ72^=(90^-18^)=18^, so sin⁡18∘sin⁡18∘=1\dfrac{\sin18^\circ}{\sin18^\circ}=118^/18^=1.

Second term. tan⁡80∘=tan⁡(90∘−10∘)=cot⁡10∘\tan80^\circ=\tan(90^\circ-10^\circ)=\cot10^\circ80^=(90^-10^)=10^, so
tan⁡10∘tan⁡80∘=tan⁡10∘cot⁡10∘=1.\tan10^\circ\tan80^\circ=\tan10^\circ\cot10^\circ=1.10^80^=10^10^=1.
Also tan⁡30∘=13\tan30^\circ=\dfrac{1}{\sqrt3}30^=1/3, so tan⁡10∘tan⁡30∘tan⁡80∘=1×13=13\tan10^\circ\tan30^\circ\tan80^\circ=1\times\dfrac{1}{\sqrt3}=\dfrac{1}{\sqrt3}10^30^80^=1×1/3=1/3.

Thus 3(13)=1\sqrt3\left(\dfrac{1}{\sqrt3}\right)=13(1/3)=1.

Total. 1+1=21+1=21+1=2.

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Q12Long AnswerHOTS5 marks

If AAA, BBB and CCC are the interior angles of a triangle ABCABCABC, prove that tan⁡(A+B2)=cot⁡(C2)\tan\left(\dfrac{A+B}{2}\right)=\cot\left(\dfrac{C}{2}\right)(A+B/2)=(C/2) and hence evaluate tan⁡(A+B2)\tan\left(\dfrac{A+B}{2}\right)(A+B/2) when C=60∘C=60^\circC=60^.

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In △ABC\triangle ABCABC, A+B+C=180∘A+B+C=180^\circA+B+C=180^.

⇒A+B=180∘−C\Rightarrow A+B=180^\circ-CA+B=180^-C

⇒A+B2=90∘−C2.\Rightarrow \dfrac{A+B}{2}=90^\circ-\dfrac{C}{2}.A+B/2=90^-C/2.

Taking tan⁡\tan of both sides,
tan⁡(A+B2)=tan⁡(90∘−C2)=cot⁡(C2).\tan\left(\dfrac{A+B}{2}\right)=\tan\left(90^\circ-\dfrac{C}{2}\right)=\cot\left(\dfrac{C}{2}\right).(A+B/2)=(90^-C/2)=(C/2).
Hence proved.

When C=60∘C=60^\circC=60^: C2=30∘\dfrac{C}{2}=30^\circC/2=30^, so
tan⁡(A+B2)=cot⁡30∘=3.\tan\left(\dfrac{A+B}{2}\right)=\cot30^\circ=\sqrt3.(A+B/2)=30^=3.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A student is simplifying trigonometric expressions using complementary angles. She writes down the identities sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ-\theta)=\cos\theta(90^-)= and tan⁡(90∘−θ)=cot⁡θ\tan(90^\circ-\theta)=\cot\theta(90^-)= and uses them to answer the following.

(i) Simplify cos⁡58∘sin⁡32∘\dfrac{\cos58^\circ}{\sin32^\circ}58^/32^.

(ii) If sin⁡3A=cos⁡(A−26∘)\sin3A=\cos(A-26^\circ)3A=(A-26^) with 3A3A3A acute, find AAA.

(iii) Evaluate cos⁡225∘+cos⁡265∘\cos^2 25^\circ+\cos^2 65^\circ^2 25^+^2 65^.

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(i) cos⁡58∘=cos⁡(90∘−32∘)=sin⁡32∘\cos58^\circ=\cos(90^\circ-32^\circ)=\sin32^\circ58^=(90^-32^)=32^, so cos⁡58∘sin⁡32∘=sin⁡32∘sin⁡32∘=1\dfrac{\cos58^\circ}{\sin32^\circ}=\dfrac{\sin32^\circ}{\sin32^\circ}=158^/32^=32^/32^=1.

(ii) cos⁡(A−26∘)=sin⁡(90∘−(A−26∘))=sin⁡(116∘−A)\cos(A-26^\circ)=\sin\big(90^\circ-(A-26^\circ)\big)=\sin(116^\circ-A)(A-26^)=(90^-(A-26^))=(116^-A).
So sin⁡3A=sin⁡(116∘−A)⇒3A=116∘−A⇒4A=116∘⇒A=29∘\sin3A=\sin(116^\circ-A)\Rightarrow 3A=116^\circ-A\Rightarrow 4A=116^\circ\Rightarrow A=29^\circ3A=(116^-A) 3A=116^-A 4A=116^ A=29^.

(iii) cos⁡65∘=cos⁡(90∘−25∘)=sin⁡25∘\cos65^\circ=\cos(90^\circ-25^\circ)=\sin25^\circ65^=(90^-25^)=25^, so
cos⁡225∘+cos⁡265∘=cos⁡225∘+sin⁡225∘=1.\cos^2 25^\circ+\cos^2 65^\circ=\cos^2 25^\circ+\sin^2 25^\circ=1.^2 25^+^2 65^=^2 25^+^2 25^=1.

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All ICSE Class 9 Maths Chapters

Rational and Irrational NumbersCompound Interest (Without Using Formula)Compound Interest (Using Formula)ExpansionsFactorisationSimultaneous (Linear) EquationsIndices (Exponents)LogarithmsTriangles (Congruency in Triangles)Isosceles TrianglesInequalitiesMid-point and Its Converse (Including Intercept Theorem)Pythagoras TheoremRectilinear Figures (Quadrilaterals)Construction of PolygonsArea TheoremsCircleStatistics (Classification of Data, Tabulation)Mean and MedianArea and Perimeter of Plane FiguresSolids (Surface Area and Volume of 3-D Solids)Trigonometrical RatiosTrigonometrical Ratios of Standard AnglesSolution of Right TrianglesComplementary AnglesCo-ordinate GeometryGraphical Solution (Simultaneous Linear Equations, Graphically)Distance Formula

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  • Do these Complementary Angles questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 9 Maths, so nothing here is outside the current course.
  • How should I practise the Complementary Angles important questions?
    Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.
  • What types of questions are covered for Complementary Angles?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the ICSE paper is covered.

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