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Complementary Angles Class 9 ICSE Important Questions with Answers

The 13 questions below are Complementary Angles Class 9 ICSE important questions, each with a full step-by-step answer: extra questions beyond the textbook exercises, covering 5 topics of ICSE Class 9 Maths in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 17 October 2026

Last updated 25 October 2026

13
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Topics
3
Key concepts
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With answers
Quick answer

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

Syllabus reference (2026–27): ICSE 2028 Mathematics syllabus (PDF) · CISCE (Council for the Indian School Certificate Examinations)

About Complementary Angles

Inside 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 Asin (90°-A)=cos A, cos⁡(90∘−A)=sin⁡A\cos(90^\circ-A)=\sin Acos (90°-A)=sin A, tan⁡(90∘−A)=cot⁡A\tan(90^\circ-A)=\cot Atan (90°-A)=cot A, cot⁡(90∘−A)=tan⁡A\cot(90^\circ-A)=\tan Acot (90°-A)=tan A, sec⁡(90∘−A)=csc⁡A\sec(90^\circ-A)=\csc Asec (90°-A)=csc A and csc⁡(90∘−A)=sec⁡A\csc(90^\circ-A)=\sec Acsc (90°-A)=sec 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 Asin (90°-A)=cos A, cos⁡(90∘−A)=sin⁡A\cos(90^\circ-A)=\sin Acos (90°-A)=sin A, tan⁡(90∘−A)=cot⁡A\tan(90^\circ-A)=\cot Atan (90°-A)=cot A, cot⁡(90∘−A)=tan⁡A\cot(90^\circ-A)=\tan Acot (90°-A)=tan A, sec⁡(90∘−A)=csc⁡A\sec(90^\circ-A)=\csc Asec (90°-A)=csc A, csc⁡(90∘−A)=sec⁡A\csc(90^\circ-A)=\sec Acsc (90°-A)=sec A.

Key ratio

sin⁡(90∘−A)cos⁡A=1\dfrac{\sin(90^\circ-A)}{\cos A}=1(sin (90°-A))/(cos A)=1 and cos⁡(90∘−A)sin⁡A=1\dfrac{\cos(90^\circ-A)}{\sin A}=1(cos (90°-A))/(sin 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 Bsin A=cos 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^\circtan A=cot B⇒ A+B=90° and sec⁡A=csc⁡B⇒A+B=90∘\sec A=\csc B\Rightarrow A+B=90^\circsec A=csc B⇒ A+B=90° (for acute angles).

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Complementary Angles Questions and Answers

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MCQs with answers (1 mark)

Q1MCQEasy1 mark

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

  1. (a)

    000

  2. (b)

    111

  3. (c)

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

  4. (d)

    3\sqrt3√3

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Answer: (b) 111.

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

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Q2MCQEasy1 mark

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

  1. (a)

    sin⁡A\sin Asin A

  2. (b)

    cos⁡A\cos Acos A

  3. (c)

    tan⁡A\tan Atan A

  4. (d)

    sec⁡A\sec Asec A

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Answer: (a) sin⁡A\sin Asin A.

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

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Q3MCQModerate1 mark

If sec⁡4A=csc⁡(A−20∘)\sec4A=\csc(A-20^\circ)sec 4A=csc (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)sec 4A=csc (90°-4A), so csc⁡(90∘−4A)=csc⁡(A−20∘)\csc(90^\circ-4A)=\csc(A-20^\circ)csc (90°-4A)=csc (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^\circtan 48° tan 23° tan 42° tan 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=1tan 48° tan 42°=tan 48° cot 48°=1 and tan⁡23∘tan⁡67∘=tan⁡23∘cot⁡23∘=1\tan23^\circ\tan67^\circ=\tan23^\circ\cot23^\circ=1tan 23° tan 67°=tan 23° cot 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}=1(tan 20°)/(cot 70°)=1.

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

  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^\circcot 70°=cot (90°-20°)=tan 20°, so tan⁡20∘tan⁡20∘=1\dfrac{\tan20^\circ}{\tan20^\circ}=1(tan 20°)/(tan 20°)=1. R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortModerate2 marks

AAA and BBB are complementary angles and sin⁡A=35\sin A=\dfrac35sin A=3/5. Find cos⁡B\cos Bcos B and tan⁡B\tan Btan B.

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B=90∘−AB=90^\circ-AB=90°-A, so cos⁡B=cos⁡(90∘−A)=sin⁡A=35\cos B=\cos(90^\circ-A)=\sin A=\dfrac35cos B=cos (90°-A)=sin A=3/5.

From sin⁡A=35\sin A=\dfrac35sin A=3/5: opposite =3=3=3, hypotenuse =5=5=5, adjacent =4=4=4, so cot⁡A=43\cot A=\dfrac43cot A=4/3. Then tan⁡B=tan⁡(90∘−A)=cot⁡A=43\tan B=\tan(90^\circ-A)=\cot A=\dfrac43tan B=tan (90°-A)=cot A=4/3.

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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^\circsec 70° sin 20°+cos 20° csc 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}sec 70°=sec (90°-20°)=csc 20°=1/(sin 20°), so sec⁡70∘sin⁡20∘=1\sec70^\circ\sin20^\circ=1sec 70° sin 20°=1.

csc⁡70∘=csc⁡(90∘−20∘)=sec⁡20∘=1cos⁡20∘\csc70^\circ=\csc(90^\circ-20^\circ)=\sec20^\circ=\dfrac{1}{\cos20^\circ}csc 70°=csc (90°-20°)=sec 20°=1/(cos 20°), so cos⁡20∘csc⁡70∘=1\cos20^\circ\csc70^\circ=1cos 20° csc 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^\circcos 75°+csc 80°+tan 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°-θ to bring each angle below 45∘45^\circ45°.

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

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

tan⁡55∘=tan⁡(90∘−35∘)=cot⁡35∘\tan55^\circ=\tan(90^\circ-35^\circ)=\cot35^\circtan 55°=tan (90°-35°)=cot 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.∴ cos 75°+csc 80°+tan 55°=sin 15°+sec 10°+cot 35°.

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Q9Short AnswerModerate3 marks

If tan⁡2A=cot⁡(A−18∘)\tan2A=\cot(A-18^\circ)tan 2A=cot (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)cot (A-18°)=tan (90°-(A-18°))=tan (108°-A).

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

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

⇒3A=108∘\Rightarrow 3A=108^\circ⇒ 3A=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(sin (90°-θ)cos (90°-θ))/(tan (90°-θ))=sin²θ.

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Convert each complementary ratio.

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

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==cos θ · sin θ(cos θ)/(sin θ)=cos θ sin θ × (sin θ)/(cos θ)=sin²θ= 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)(sin 18°)/(cos 72°)+√3(tan 10° tan 30° tan 80° ).

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

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

Thus 3(13)=1\sqrt3\left(\dfrac{1}{\sqrt3}\right)=1√3(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)tan ((A+B)/2)=cot (C/2) and hence evaluate tan⁡(A+B2)\tan\left(\dfrac{A+B}{2}\right)tan ((A+B)/2) when C=60∘C=60^\circC=60°.

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

⇒A+B=180∘−C\Rightarrow A+B=180^\circ-C⇒ A+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⁡\tantan 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).tan ((A+B)/2)=tan (90°-C/2)=cot (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.tan ((A+B)/2)=cot 30°=√3.

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Case study 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\thetasin (90°-θ)=cos θ and tan⁡(90∘−θ)=cot⁡θ\tan(90^\circ-\theta)=\cot\thetatan (90°-θ)=cot θ and uses them to answer the following.

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

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

(iii) Evaluate cos⁡225∘+cos⁡265∘\cos^2 25^\circ+\cos^2 65^\circcos² 25°+cos² 65°.

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(i) cos⁡58∘=cos⁡(90∘−32∘)=sin⁡32∘\cos58^\circ=\cos(90^\circ-32^\circ)=\sin32^\circcos 58°=cos (90°-32°)=sin 32°, so cos⁡58∘sin⁡32∘=sin⁡32∘sin⁡32∘=1\dfrac{\cos58^\circ}{\sin32^\circ}=\dfrac{\sin32^\circ}{\sin32^\circ}=1(cos 58°)/(sin 32°)=(sin 32°)/(sin 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)cos (A-26°)=sin (90°-(A-26°))=sin (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^\circsin 3A=sin (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^\circcos 65°=cos (90°-25°)=sin 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.cos² 25°+cos² 65°=cos² 25°+sin² 25°=1.

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Continues in Class 10: Trigonometrical Identities Class 10 ICSE questions

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

Frequently asked questions

  • What are the complementary angle formulas in trigonometry?
    sin⁡(90∘−A)=cos⁡A\sin(90^\circ-A)=\cos Asin (90°-A)=cos A, cos⁡(90∘−A)=sin⁡A\cos(90^\circ-A)=\sin Acos (90°-A)=sin A, tan⁡(90∘−A)=cot⁡A\tan(90^\circ-A)=\cot Atan (90°-A)=cot A, cot⁡(90∘−A)=tan⁡A\cot(90^\circ-A)=\tan Acot (90°-A)=tan A, sec⁡(90∘−A)=csc⁡A\sec(90^\circ-A)=\csc Asec (90°-A)=csc A and csc⁡(90∘−A)=sec⁡A\csc(90^\circ-A)=\sec Acsc (90°-A)=sec A. So sin⁡35∘cos⁡55∘=1\tfrac{\sin35^\circ}{\cos55^\circ}=1(sin 35°)/(cos 55°)=1. Each ratio of an angle equals the co-ratio of its complement, which is where the 'co' in cosine, cotangent and cosecant comes from. Questions test this by pairing angles that add up to 90circ90^circ90^circ.
  • How do you solve sec 4A = cosec(A − 20°)?
    Write sec⁡4A=csc⁡(90∘−4A)\sec4A=\csc(90^\circ-4A)sec 4A=csc (90°-4A), so 90∘−4A=A−20∘90^\circ-4A=A-20^\circ90°-4A=A-20°, giving 5A=110∘5A=110^\circ5A=110° and A=22∘A=22^\circA=22°. In general, if a ratio of one angle equals the co-ratio of another, the two angles add to 90∘90^\circ90°.
  • What is tan 48° tan 23° tan 42° tan 67°?
    Pair complementary angles: tan⁡48∘tan⁡42∘=tan⁡48∘cot⁡48∘=1\tan48^\circ\tan42^\circ=\tan48^\circ\cot48^\circ=1tan 48° tan 42°=tan 48° cot 48°=1 and tan⁡23∘tan⁡67∘=1\tan23^\circ\tan67^\circ=1tan 23° tan 67°=1, so the product is 111. Looking for pairs adding to 90∘90^\circ90° is the whole trick. The same pairing works for products of sines and cosecants.
  • If A and B are complementary and sin A = 3/5, what are cos B and tan B?
    cos⁡B=cos⁡(90∘−A)=sin⁡A=35\cos B=\cos(90^\circ-A)=\sin A=\tfrac35cos B=cos (90°-A)=sin A=3/5. Then sin⁡B=cos⁡A=45\sin B=\cos A=\tfrac45sin B=cos A=4/5, so tan⁡B=4/53/5=43\tan B=\tfrac{4/5}{3/5}=\tfrac43tan B=4/5/3/5=4/3. Here the 3,4,53, 4, 53, 4, 5 triangle gives every ratio at once. Draw it with AAA and BBB as the two acute angles and read off the sides opposite and adjacent to BBB; this picture makes the complementary relations obvious.
  • Which ICSE Class 9 chapter comes after Complementary Angles?
    Co-ordinate Geometry (Chapter 26) follows Complementary Angles (Chapter 25), which closes the Class 9 trigonometry chapters. In Class 10 the topic continues with Trigonometrical Identities (Chapter 21).
  • How do I evaluate complementary angle expressions without tables?
    Rewrite one angle of each pair as 90∘90^\circ90° minus the other, convert to the co-ratio, and cancel or multiply to 111. For sin⁡18∘cos⁡72∘+3(tan⁡10∘tan⁡30∘tan⁡80∘)\tfrac{\sin18^\circ}{\cos72^\circ}+\sqrt3(\tan10^\circ\tan30^\circ\tan80^\circ)(sin 18°)/(cos 72°)+√3(tan 10° tan 30° tan 80°) this gives 1+3×13=21+\sqrt3\times\tfrac1{\sqrt3}=21+√3× 1/√3=2.
  • Can I download Complementary Angles ICSE Class 9 questions?
    Yes. Enter your email in the free PDF box on this page and all 13 complementary-angle questions open with worked answers as a printable worksheet.

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