If \(\cos A=\dfrac{4}{5}\), then the value of \(\tan A\) is (A) \(\dfrac{3}{5}\) (B) \(\dfrac{3}{4}\) (C) \(\dfrac{

Correct option: (B) \(\dfrac{3}{4}\). Concept used. In a right triangle, \(\cos A=\dfrac{\text{adjacent}}{\text{hypotenuse}}\). The third side comes from the Pythagoras theorem, and then \(\tan A=\dfrac{\text{opposite}}{\text{adjacent}}\). Read off \(\cos A=\dfrac45\), so adjacen

If \(\sin A=\dfrac{1}{2}\), then the value of \(\cot A\) is (A) \(\sqrt{3}\) (B) \(\dfrac{1}{\sqrt{3}}\) (C) \(\dfr

Correct option: (A) \(\sqrt{3}\). Concept used. The value \(\sin A=\tfrac12\) matches a standard angle. Once the angle is known, every other ratio is a table value. \(\sin A=\dfrac12\) holds only for \(A=30^\circ\) among acute angles. From the standard table, \(\cot 30^\circ=\sqr

The value of the expression \(\big[\csc(75^\circ+\theta)-\sec(15^\circ-\theta)-\tan(55^\circ+\theta)+\cot(35^\circ-\theta)\big]\) is (A) \(-1\) (B)&nb

Correct option: (B) \(0\). Concept used. The complementary-angle rules \(\csc(90^\circ-A)=\sec A\) and \(\tan(90^\circ-A)=\cot A\). Pair the terms whose angles add to \(90^\circ\). The angles \((75^\circ+\theta)\) and \((15^\circ-\theta)\) add to \(90^\circ\), so [] \(\csc(75^\ci

Given that \(\sin\theta=\dfrac{a}{b}\), then \(\cos\theta\) is equal to (A) \(\dfrac{b}{\sqrt{b^2-a^2}}\) (B) \(\dfrac{b}{a}\)

Correct option: (C) \(\dfrac{\sqrt{b^2-a^2}}{b}\). Concept used. The identity \(\sin^2\theta+\cos^2\theta=1\), which lets you find one ratio from the other. Start from the identity and isolate the cosine: [] \(\cos^2\theta=1-\sin^2\theta\). Substitute \(\sin\theta=\dfrac{a}{b}\):

If \(\cos(\alpha+\beta)=0\), then \(\sin(\alpha-\beta)\) can be reduced to (A) \(\cos\beta\) (B) \(\cos 2\beta\) (C)&nbs

Correct option: (B) \(\cos 2\beta\). Concept used. \(\cos\,(\text{angle})=0\) means the angle is \(90^\circ\), plus the complementary rule \(\sin(90^\circ-x)=\cos x\). \(\cos(\alpha+\beta)=0\) forces \(\alpha+\beta=90^\circ\), so \(\alpha=90^\circ-\beta\). Substitute into \(\sin(

The value of \((\tan 1^\circ\,\tan 2^\circ\,\tan 3^\circ\ldots\tan 89^\circ)\) is (A) \(0\) (B) \(1\) (C) \(2\) &nb

Correct option: (B) \(1\). Concept used. The complementary rule \(\tan(90^\circ-A)=\cot A\), together with \(\tan A\cdot\cot A=1\). Pair the first and last factors: \(\tan 1^\circ\) with \(\tan 89^\circ\). Since \(89^\circ=90^\circ-1^\circ\), [] \(\tan 89^\circ=\cot 1^\circ\), so

If \(\cos 9\alpha=\sin\alpha\) and \(9\alpha (A) \(\dfrac{1}{\sqrt{3}}\) (B) \(\sqrt{3}\) (C) \(1\) &nb

Correct option: (C) \(1\). Concept used. The complementary rule \(\sin\alpha=\cos(90^\circ-\alpha)\), which converts a sine to a cosine so both sides can be compared. Rewrite the right side as a cosine: [] \(\sin\alpha=\cos(90^\circ-\alpha)\). The equation becomes \(\cos 9\alpha=

If \(\sin A+\sin^2 A=1\), then the value of the expression \((\cos^2 A+\cos^4 A)\) is (A) \(1\) (B) \(\dfrac{1}{2}\) (C)

Correct option: (A) \(1\). Concept used. The identity \(\sin^2 A+\cos^2 A=1\), used to turn the given condition into a statement about \(\cos^2 A\). From the condition, \(\sin A=1-\sin^2 A\). But \(1-\sin^2 A=\cos^2 A\), so [] \(\sin A=\cos^2 A\). Now rewrite the target. Replace

Given that \(\sin\alpha=\dfrac{1}{2}\) and \(\cos\beta=\dfrac{1}{2}\), then the value of \((\alpha+\beta)\) is (A) \(0^\circ\) (B) \(30^\circ\) &

Correct option: (D) \(90^\circ\). Concept used. The standard-angle table: read the angle whose sine (or cosine) equals \(\tfrac12\). \(\sin\alpha=\dfrac12\) gives \(\alpha=30^\circ\). \(\cos\beta=\dfrac12\) gives \(\beta=60^\circ\). Add them: [] \(\alpha+\beta=30^\circ+60^\circ=9

The value of the expression \(\left[\dfrac{\sin^2 22^\circ+\sin^2 68^\circ}{\cos^2 22^\circ+\cos^2 68^\circ}+\sin^2 63^\circ+\cos 63^\circ\sin 27^\circ\right]\) is (A) \(3\) &

Correct option: (B) \(2\). Concept used. Complementary rules \(\sin 68^\circ=\cos 22^\circ\), \(\cos 68^\circ=\sin 22^\circ\), \(\sin 27^\circ=\cos 63^\circ\), plus \(\sin^2\theta+\cos^2\theta=1\). In the numerator, \(\sin 68^\circ=\cos 22^\circ\), so [] \(\sin^2 22^\circ+\sin^2

If \(4\tan\theta=3\), then \(\left(\dfrac{4\sin\theta-\cos\theta}{4\sin\theta+\cos\theta}\right)\) is equal to (A) \(\dfrac{2}{3}\) (B) \(\dfrac{

Correct option: (C) \(\dfrac{1}{2}\). Concept used. Dividing numerator and denominator by \(\cos\theta\) turns every term into \(\tan\theta\), which is given. From \(4\tan\theta=3\) we get \(\tan\theta=\dfrac34\). Divide top and bottom of the expression by \(\cos\theta\): [] \(\d

If \(\sin\theta-\cos\theta=0\), then the value of \((\sin^4\theta+\cos^4\theta)\) is (A) \(1\) (B) \(\dfrac{3}{4}\) (C)&

Correct option: (C) \(\dfrac{1}{2}\). Concept used. \(\sin\theta=\cos\theta\) means \(\tan\theta=1\), so \(\theta=45^\circ\); then use the standard values. \(\sin\theta-\cos\theta=0\) gives \(\sin\theta=\cos\theta\), i.e. \(\tan\theta=1\), so \(\theta=45^\circ\). At \(45^\circ\),

\(\sin(45^\circ+\theta)-\cos(45^\circ-\theta)\) is equal to (A) \(2\cos\theta\) (B) \(0\) (C) \(2\sin\theta\) &nbsp

Correct option: (B) \(0\). Concept used. The complementary rule \(\cos(90^\circ-x)=\sin x\). Rewrite the cosine term using \(45^\circ-\theta=90^\circ-(45^\circ+\theta)\): [] \(\cos(45^\circ-\theta)=\cos\big(90^\circ-(45^\circ+\theta)\big)=\sin(45^\circ+\theta)\). Substitute back:

\(\dfrac{\tan 47^\circ}{\cot 43^\circ}=1\). State true or false and justify.

Verdict: True. The ratio equals \(1\). Concept used. The complementary-angle rule \(\cot 43^\circ=\tan(90^\circ-43^\circ)=\tan 47^\circ\). Convert the cotangent to a tangent: [] \(\cot 43^\circ=\tan(90^\circ-43^\circ)=\tan 47^\circ\). Substitute into the fraction: [] \(\dfrac{\ta

The value of the expression \((\cos^2 23^\circ-\sin^2 67^\circ)\) is positive. State true or false and justify.

Verdict: False. The expression equals \(0\), which is not positive. Concept used. The complementary rule \(\sin 67^\circ=\cos 23^\circ\) (since \(67^\circ=90^\circ-23^\circ\)). Rewrite the sine using the complement: [] \(\sin 67^\circ=\sin(90^\circ-23^\circ)=\cos 23^\circ\). So \

The value of the expression \((\sin 80^\circ-\cos 80^\circ)\) is negative. State true or false and justify.

Verdict: False. The expression is positive. Concept used. As the acute angle grows, \(\sin\theta\) increases and \(\cos\theta\) decreases; also \(\cos 80^\circ=\sin 10^\circ\). Write the cosine as a sine of the complement: [] \(\cos 80^\circ=\sin(90^\circ-80^\circ)=\sin 10^\circ\

\((1-\cos^2\theta)\sec^2\theta=\tan\theta\). State true or false and justify.

Verdict: False. The left side simplifies to \(\tan^2\theta\), not \(\tan\theta\). Concept used. The identities \(1-\cos^2\theta=\sin^2\theta\) and \(\sec\theta=\dfrac{1}{\cos\theta}\), with \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\). Replace \(1-\cos^2\theta\) by \(\sin^2\thet

If \(\cos A+\cos^2 A=1\), then \(\sin^2 A+\sin^4 A=1\). State true or false and justify.

Verdict: True. The second equation does follow. Concept used. The identity \(\sin^2 A=1-\cos^2 A\), used to convert the given condition. From the condition, \(\cos A=1-\cos^2 A\). But \(1-\cos^2 A=\sin^2 A\), so [] \(\cos A=\sin^2 A\). Now simplify the target. First, \(\sin^4 A=(

\((\tan\theta+2)(2\tan\theta+1)=5\tan\theta+\sec^2\theta\). State true or false and justify.

Verdict: False. The two sides differ by \(1\). Concept used. Ordinary algebraic expansion, plus the identity \(1+\tan^2\theta=\sec^2\theta\). Expand the left side: [] \((\tan\theta+2)(2\tan\theta+1)=2\tan^2\theta+\tan\theta+4\tan\theta+2\) [] \(=2\tan^2\theta+5\tan\theta+2\). Now

The value of \(2\sin\theta\) can be \(a+\dfrac{1}{a}\), where \(a\) is a positive number, and \(a\ne 1\). State true or false and justify.

Verdict: False. No such \(\theta\) exists. Concept used. The bound \(-1\le\sin\theta\le 1\), so \(-2\le 2\sin\theta\le 2\), together with the AM-GM fact \(a+\dfrac1a\ge 2\) for \(a>0\). For any angle, \(\sin\theta\le 1\), so [] \(2\sin\theta\le 2\). For a positive \(a\), the AM-G

Prove that \(\dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}=2\csc\theta\).

Concept used. Add the two fractions over a common denominator, then apply \(\sin^2\theta+\cos^2\theta=1\) and \(\csc\theta=\dfrac{1}{\sin\theta}\). Take the common denominator \(\sin\theta(1+\cos\theta)\): [] \(\text{L.H.S.}=\dfrac{\sin^2\theta+(1+\cos\theta)^2}{\sin\theta(1+\cos

Prove that \(\dfrac{\tan A}{1+\sec A}-\dfrac{\tan A}{1-\sec A}=2\csc A\).

Concept used. Subtract over a common denominator, then use \(1-\sec^2 A=-\tan^2 A\) and the ratio forms \(\tan A=\dfrac{\sin A}{\cos A}\), \(\sec A=\dfrac{1}{\cos A}\). Common denominator is \((1+\sec A)(1-\sec A)=1-\sec^2 A\): [] \(\text{L.H.S.}=\dfrac{\tan A(1-\sec A)-\tan A(1+

If \(\tan A=\dfrac{3}{4}\), then show that \(\sin A\cos A=\dfrac{12}{25}\).

Concept used. Build the right triangle from \(\tan A\), find the hypotenuse by Pythagoras, then read \(\sin A\) and \(\cos A\). \(\tan A=\dfrac34\) means opposite \(=3\), adjacent \(=4\). Hypotenuse by Pythagoras: [] \(\text{hyp}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\). So \(\si

Prove that \((\sin\alpha+\cos\alpha)(\tan\alpha+\cot\alpha)=\sec\alpha+\csc\alpha\).

Concept used. Convert \(\tan\alpha\) and \(\cot\alpha\) to sine and cosine, simplify \(\tan\alpha+\cot\alpha\) using \(\sin^2\alpha+\cos^2\alpha=1\). Simplify the second bracket: [] \(\tan\alpha+\cot\alpha=\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{\sin^

Prove that \((\sqrt{3}+1)(3-\cot 30^\circ)=\tan^3 60^\circ-2\sin 60^\circ\).

Concept used. Substitute the standard values \(\cot 30^\circ=\sqrt3\), \(\tan 60^\circ=\sqrt3\) and \(\sin 60^\circ=\dfrac{\sqrt3}{2}\), then simplify both sides. Left side, with \(\cot 30^\circ=\sqrt3\): [] \((\sqrt3+1)(3-\sqrt3)\). Expand: [] \(=3\sqrt3-\sqrt3\cdot\sqrt3+3-\sqr

Prove that \(1+\dfrac{\cot^2\alpha}{1+\csc\alpha}=\csc\alpha\).

Concept used. Replace \(\cot^2\alpha\) using \(\cot^2\alpha=\csc^2\alpha-1\), then factorise the difference of squares. Substitute \(\cot^2\alpha=\csc^2\alpha-1\) in the second term: [] \(\dfrac{\cot^2\alpha}{1+\csc\alpha}=\dfrac{\csc^2\alpha-1}{1+\csc\alpha}\). Factorise the num

Prove that \(\tan\theta+\tan(90^\circ-\theta)=\sec\theta\,\sec(90^\circ-\theta)\).

Concept used. Complementary rules \(\tan(90^\circ-\theta)=\cot\theta\) and \(\sec(90^\circ-\theta)=\csc\theta\), then convert to sine and cosine. Replace the complementary terms: [] \(\tan(90^\circ-\theta)=\cot\theta,\qquad \sec(90^\circ-\theta)=\csc\theta\). Left side becomes: [

If \(\sqrt{3}\tan\theta=1\), then find the value of \(\sin^2\theta-\cos^2\theta\).

Concept used. Solve for \(\theta\) from \(\tan\theta=\dfrac{1}{\sqrt3}\), then substitute standard values. From \(\sqrt3\tan\theta=1\): [] \(\tan\theta=\dfrac{1}{\sqrt3}\), so \(\theta=30^\circ\). Standard values: \(\sin 30^\circ=\dfrac12\), \(\cos 30^\circ=\dfrac{\sqrt3}{2}\). S

Simplify \((1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)\).

Concept used. The identities \(1+\tan^2\theta=\sec^2\theta\) and the difference of squares \((1-\sin\theta)(1+\sin\theta)=1-\sin^2\theta\). Multiply the last two brackets first: [] \((1-\sin\theta)(1+\sin\theta)=1-\sin^2\theta=\cos^2\theta\). Replace the first bracket using the i

If \(2\sin^2\theta-\cos^2\theta=2\), then find the value of \(\theta\).

Concept used. Replace \(\cos^2\theta\) by \(1-\sin^2\theta\) to get an equation in \(\sin\theta\) alone. Substitute \(\cos^2\theta=1-\sin^2\theta\): [] \(2\sin^2\theta-(1-\sin^2\theta)=2\). Open the bracket and collect: [] \(2\sin^2\theta-1+\sin^2\theta=2\) [] \(3\sin^2\theta-1=2

Show that \(\dfrac{\cos^2(45^\circ+\theta)+\cos^2(45^\circ-\theta)}{\tan(60^\circ+\theta)\tan(30^\circ-\theta)}=1\).

Concept used. For the numerator, \(\cos(45^\circ-\theta)=\sin(45^\circ+\theta)\) gives \(\sin^2+\cos^2=1\). For the denominator, \(\tan(30^\circ-\theta)=\cot(60^\circ+\theta)\) gives a product of \(1\). Numerator: since \((45^\circ-\theta)=90^\circ-(45^\circ+\theta)\), [] \(\cos(

Show that \(\tan^4\theta+\tan^2\theta=\sec^4\theta-\sec^2\theta\).

Concept used. Factor the left side, then use \(1+\tan^2\theta=\sec^2\theta\) twice. Take \(\tan^2\theta\) common on the left: [] \(\tan^4\theta+\tan^2\theta=\tan^2\theta(\tan^2\theta+1)\). Replace \(\tan^2\theta+1\) by \(\sec^2\theta\): [] \(=\tan^2\theta\,\sec^2\theta\). Now rep

If \(\csc\theta+\cot\theta=p\), then prove that \(\cos\theta=\dfrac{p^2-1}{p^2+1}\).

Concept used. Square the given relation, use \(\csc^2\theta-\cot^2\theta=1\), and write \(\csc\theta\), \(\cot\theta\) in terms of \(\cos\theta\). Write the given in sine and cosine: [] \(p=\csc\theta+\cot\theta=\dfrac{1}{\sin\theta}+\dfrac{\cos\theta}{\sin\theta}=\dfrac{1+\cos\t

Prove that \(\sqrt{\sec^2\theta+\csc^2\theta}=\tan\theta+\cot\theta\).

Concept used. Replace \(\sec^2\theta\) and \(\csc^2\theta\) by their \(\sin\)-\(\cos\) forms, combine over a common denominator, and use \(\sin^2\theta+\cos^2\theta=1\). Work on the inside of the root: [] \(\sec^2\theta+\csc^2\theta=\dfrac{1}{\cos^2\theta}+\dfrac{1}{\sin^2\theta}

If \(1+\sin^2\theta=3\sin\theta\cos\theta\), then prove that \(\tan\theta=1\) or \(\tan\theta=\dfrac{1}{2}\).

Concept used. Replace the constant \(1\) by \(\sin^2\theta+\cos^2\theta\), then divide through by \(\cos^2\theta\) to get a quadratic in \(\tan\theta\). Write \(1=\sin^2\theta+\cos^2\theta\) on the left: [] \(\sin^2\theta+\cos^2\theta+\sin^2\theta=3\sin\theta\cos\theta\) [] \(2\s

Given that \(\sin\theta+2\cos\theta=1\), then prove that \(2\sin\theta-\cos\theta=2\).

Concept used. Square the given relation, add the square of the target, and use \(\sin^2\theta+\cos^2\theta=1\) to fix the value. Consider the sum of the squares of both expressions: [] \((\sin\theta+2\cos\theta)^2+(2\sin\theta-\cos\theta)^2\). Expand each square: [] \((\sin\theta

If \(\tan\theta+\sec\theta=l\), then prove that \(\sec\theta=\dfrac{l^2+1}{2l}\).

Concept used. Use \(\sec^2\theta-\tan^2\theta=1\), which factors as \((\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1\). The given is \(\sec\theta+\tan\theta=l\). From \(\sec^2\theta-\tan^2\theta=1\): [] \((\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1\). Divide by the given \

If \(\sin\theta+\cos\theta=p\) and \(\sec\theta+\csc\theta=q\), then prove that \(q(p^2-1)=2p\).

Concept used. Expand \(p^2=(\sin\theta+\cos\theta)^2\) to get \(2\sin\theta\cos\theta\), and write \(q\) as a single fraction. Square \(p\): [] \(p^2=(\sin\theta+\cos\theta)^2=\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1+2\sin\theta\cos\theta\). So \(p^2-1=2\sin\theta\cos\th

If \(a\sin\theta+b\cos\theta=c\), then prove that \(a\cos\theta-b\sin\theta=\sqrt{a^2+b^2-c^2}\).

Concept used. Add the squares of the two expressions; the cross terms cancel and \(\sin^2\theta+\cos^2\theta=1\) leaves \(a^2+b^2\). Square both expressions and add: [] \((a\sin\theta+b\cos\theta)^2+(a\cos\theta-b\sin\theta)^2\). Expand each: [] \((a\sin\theta+b\cos\theta)^2=a^2\

Prove that \(\dfrac{1+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}=\dfrac{1-\sin\theta}{\cos\theta}\).

Concept used. Insert \(\sec^2\theta-\tan^2\theta=1\) as a hidden factor on the left, then convert to sine and cosine. In the numerator, replace the leading \(1\) by \(\sec^2\theta-\tan^2\theta\): [] \(1+\sec\theta-\tan\theta=(\sec^2\theta-\tan^2\theta)+(\sec\theta-\tan\theta)\).

NCERT Exemplar Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry

These NCERT Exemplar Class 10 Maths Chapter 8 Introduction to Trigonometry Solutions cover every problem from Exercises 8.1 to 8.4 with step-by-step working. Each answer shows how to apply trigonometric ratios, identities, and complementary angle relations. The set follows the 2026-27 CBSE syllabus.

Exemplar problems across four exercises: MCQs, true-or-false, short-answer, and long-answer questions.
Covers trigonometric ratios of standard angles (0°, 30°, 45°, 60°, 90°), complementary angles, and the Pythagorean identities.
Free PDF download plus an inline solved question bank you can open on this page.

NCERT Exemplar Class 10 Maths Chapter 8 Introduction to Trigonometry Solutions

Solved by Collegedunia: Every question here is worked out by our Mathematics faculty, cross-checked against the official NCERT Exemplar, and aligned to the 2026-27 CBSE syllabus.

Table of Contents

Exemplar Question Types at a Glance
Key Ratios, Identities & Angle Values
How These Solutions Help You
Exemplar vs Textbook: Where It Gets Harder
Common Mistakes to Avoid
Other Resources for This Chapter
All Exemplar Questions with Step-by-Step Solutions

Watch Introduction to Trigonometry Class 10 Maths Explained

Source: Magnet Brains on YouTube

Exemplar Question Types at a Glance

The Exemplar spans four exercises covering every Trigonometry idea in the syllabus. Each one targets a different level of thinking, from direct ratio recall to multi-step identity proofs.

Exercise	Type	Count	What It Tests
Exercise 8.1	MCQ	20	Recall and apply trig ratios, standard angle values, and complementary-angle relations
Exercise 8.2	True or False	10	Check whether a statement holds; needs a written reason or counterexample
Exercise 8.3	Short answer	17	Evaluate expressions, find unknown angles, simplify using identities
Exercise 8.4	Long answer	8	Prove identities and solve multi-step problems combining two or more identities

The full set has 55 problems. A smart order: use the MCQs to lock in standard angle values and ratio definitions, sharpen justification writing with the true-or-false set, build fluency with the short-answer problems, then take on the identity proofs.

Key Ratios, Identities & Angle Values

Almost every ncert exemplar class 10 maths chapter 8 problem relies on one of the three results below. Know them cold before you open the exercises. It saves time and prevents slips in the board exam.

Standard Angle Values

Ratio	0°	30°	45°	60°	90°
sin	0	12	12	32	1
cos	1	32	12	12	0
tan	0	13	1	3	undefined

Fundamental Trigonometric Identities

Pythagorean Identity 1: sin²A + cos²A = 1. This is the most frequently used identity across all Exemplar exercises. Every identity proof in Exercise 8.4 either starts here or lands here.
Pythagorean Identity 2: 1 + tan²A = sec²A. Used when the expression involves tan and sec together.
Pythagorean Identity 3: 1 + cot²A = cosec²A. Used when the expression involves cot and cosec.

Complementary Angle Relations

sin(90° - A) = cos A and cos(90° - A) = sin A. These pair up nicely to simplify sums of the form sin A + sin(90° - A).
tan(90° - A) = cot A; similarly sec(90° - A) = cosec A and cosec(90° - A) = sec A. The MCQs in Exercise 8.1 test these repeatedly.

A time-saving tip: before substituting, rewrite every ratio in terms of sin and cos. This collapses most multi-ratio expressions into a simpler form that uses only the Pythagorean identity, and it is far less error-prone under exam pressure.

How These Solutions Help You

These solutions are built for self-study in the weeks before the board exam. They do three things for you:

Show every step: the identity is written first, both sides are simplified step by step, then confirmed equal, so you can see exactly where your own working diverges.
Justify true-or-false answers fully: every verdict gives the exact counterexample or theorem behind it. This is the part most students skip and lose marks on.
Add an Expert view: each question has a second, faster method, like factoring before you substitute, or using a complementary angle relation instead of expanding the full expression.

Use them like this: attempt the question first, open Check Solution to compare step by step, then read Expert Solution. That order builds real recall, not just recognition.

Exemplar vs Textbook: Where It Gets Harder

The textbook exercises test one skill at a time: find a ratio, evaluate at a standard angle, or use one complementary relation. The Exemplar pushes the same ideas into multi-step MCQs, justification tasks, and combined-identity proofs. The table shows where the step-up happens.

Skill	NCERT Textbook	NCERT Exemplar
Trigonometric Ratios	Find sin A, cos A, tan A from a given right triangle	Determine which ratio combination simplifies to a given value; verify whether a statement about ratios is true or false with justification
Standard Angle Values	Evaluate a single expression at 30°, 45°, or 60°	Evaluate multi-ratio expressions; prove equalities using standard values; MCQs where all four options differ only in which angle is substituted
Complementary Angles	Apply sin(90°-A) = cos A to simplify one pair	Simplify sums with multiple complementary pairs; prove equivalence using both the relation and a Pythagorean identity in the same step
Trigonometric Identities	Verify that sin²A + cos²A = 1 for a specific angle	Prove multi-step identities; manipulate LHS and RHS simultaneously using all three Pythagorean identities plus reciprocal relations
Combined problems	One technique per question	Multiple identities in one problem; for example, prove an identity that requires using all three Pythagorean identities and complementary angle relations in sequence

This is why doing the Exemplar after the textbook is the standard board-prep route for Trigonometry. The textbook teaches each result one at a time; the Exemplar makes you pick the right one under MCQ pressure and in multi-step proofs.

Common Mistakes to Avoid

Across Exercises 8.1 to 8.4, these four slips cost the most marks in the board exam. Catch them before your exam.

Treating tan 90° as a large number, not undefined: tan 90° is undefined because cos 90° = 0. Some MCQ options test exactly this. A numerical value for tan 90° in a proof breaks the whole argument.
Using a complementary relation the wrong way: sin(90° - A) = cos A holds for any acute angle A. Students get this right with a single letter but misread the complement when the angle is an expression, like A = (90° - B).
Squaring both sides of a proof: you must simplify one side to match the other. Assuming LHS = RHS and squaring both sides adds extra solutions. Exercise 8.4 gives zero marks for such circular arguments.
Weak justification in true-or-false: just "True" or "False" with no counterexample or theorem earns zero marks. The rubric needs a short reason every time.

For most students, one or two of these four slips cause nearly all the lost marks. Keep a short written list of your own repeat mistakes and fix them before boards.

Other Resources for This Chapter

Pair this Exemplar set with the other Chapter 8 resources to cover the whole chapter before your board exam.

Resource	Open
Exemplar Solutions	Introduction to Trigonometry Exemplar Solutions
NCERT Solutions	Introduction to Trigonometry NCERT Solutions
Revision Notes	Introduction to Trigonometry Notes
Formula Sheet	Introduction to Trigonometry Formula Sheet
Handwritten Notes	Introduction to Trigonometry Handwritten Notes
NCERT Book PDF	Introduction to Trigonometry NCERT Book PDF
Exemplar Book PDF	Introduction to Trigonometry Exemplar Book PDF

All Exemplar Questions with Step-by-Step Solutions

I. Multiple Choice Questions (Exercise 8.1)

Q 8.1

If cos A=45, then the value of tan A is
(A) 35 (B) 34 (C) 43 (D) 53

Q 8.2

If sin A=12, then the value of cot A is
(A) √3 (B) 1√3 (C) √32 (D) 1

Q 8.3

The value of the expression [csc(75^∘+θ)-sec(15^∘-θ)-tan(55^∘+θ)+cot(35^∘-θ)] is
(A) -1 (B) 0 (C) 1 (D) 32

Q 8.4

Given that sinθ=ab, then cosθ is equal to
(A) b√b²-a² (B) ba (C) √b²-a²b (D) a√b²-a²

Q 8.5

If cos(α+β)=0, then sin(α-β) can be reduced to
(A) cosβ (B) cos 2β (C) sinα (D) sin 2α

Q 8.6

The value of (tan 1^∘ tan 2^∘ tan 3^∘89^∘) is
(A) 0 (B) 1 (C) 2 (D) 12

Q 8.7

If cos 9α=sinα and 9α<90^∘, then the value of tan 5α is
(A) 1√3 (B) √3 (C) 1 (D) 0

Q 8.8

If ABC is right angled at C, then the value of cos(A+B) is
(A) 0 (B) 1 (C) 12 (D) √32

Q 8.9

If sin A+sin² A=1, then the value of the expression (cos² A+cos⁴ A) is
(A) 1 (B) 12 (C) 2 (D) 3

Q 8.10

Given that sinα=12 and cosβ=12, then the value of (α+β) is
(A) 0^∘ (B) 30^∘ (C) 60^∘ (D) 90^∘

Q 8.11

The value of the expression [sin² 22^∘+sin² 68^∘cos² 22^∘+cos² 68^∘+sin² 63^∘+cos 63 27^∘] is
(A) 3 (B) 2 (C) 1 (D) 0

Q 8.12

If 4tanθ=3, then (4sinθ-cosθ4sinθ+cosθ) is equal to
(A) 23 (B) 13 (C) 12 (D) 34

Q 8.13

If sinθ-cosθ=0, then the value of (sin⁴θ+cos⁴θ) is
(A) 1 (B) 34 (C) 12 (D) 14

Q 8.14

sin(45^∘+θ)-cos(45^∘-θ) is equal to
(A) 2cosθ (B) 0 (C) 2sinθ (D) 1

NCERT exemplar Class 12 Mathematics Chapter 8 Introduction to Trigonometry

Introduction to Trigonometry NCERT Exemplar

All 8 questions with collapsible Solution and Expert Solution. Tap a button to reveal the working.

II. True / False with Reasoning (Exercise 8.2)

Q 8.1

tan 47^∘cot 43^∘=1. State true or false and justify.

Q 8.2

The value of the expression (cos² 23^∘-sin² 67^∘) is positive. State true or false and justify.

Q 8.3

The value of the expression (sin 80^∘-cos 80^∘) is negative. State true or false and justify.

Q 8.4

(1-cos²θ)sec²θ=tanθ. State true or false and justify.

Q 8.5

If cos A+cos² A=1, then sin² A+sin⁴ A=1. State true or false and justify.

Q 8.6

(tanθ+2)(2tanθ+1)=5tanθ+sec²θ. State true or false and justify.

Q 8.7

The value of 2sinθ can be a+1a, where a is a positive number, and a≠ 1. State true or false and justify.

Q 8.8

cosθ=a²+b²2ab, where a and b are two distinct numbers such that ab>0. State true or false and justify.

NCERT exemplar Class 12 Mathematics Chapter 8 Introduction to Trigonometry

Introduction to Trigonometry NCERT Exemplar

All 12 questions with collapsible Solution and Expert Solution. Tap a button to reveal the working.

III. Short Answer Questions (Exercise 8.3)

Q 8.1

Prove that sinθ1+cosθ+1+cosθsinθ=2cscθ.

Q 8.2

Prove that tan A1+sec A-tan A1-sec A=2csc A.

Q 8.3

If tan A=34, then show that sin Acos A=1225.

Q 8.4

Prove that (sinα+cosα)(tanα+cotα)=secα+cscα.

Q 8.5

Prove that (√3+1)(3-cot 30^∘)=tan³ 60^∘-2sin 60^∘.

Q 8.6

Prove that 1+cot²α1+cscα=cscα.

Q 8.7

Prove that tanθ+tan(90^∘-θ)=secθ sec(90^∘-θ).

Q 8.8

If √3tanθ=1, then find the value of sin²θ-cos²θ.

Q 8.9

Simplify (1+tan²θ)(1-sinθ)(1+sinθ).

Q 8.10

If 2sin²θ-cos²θ=2, then find the value of θ.

Q 8.11

Show that cos²(45^∘+θ)+cos²(45^∘-θ)tan(60^∘+θ)tan(30^∘-θ)=1.

Q 8.12

Show that tan⁴θ+tan²θ=sec⁴θ-sec²θ.

NCERT exemplar Class 12 Mathematics Chapter 8 Introduction to Trigonometry

Introduction to Trigonometry NCERT Exemplar

All 8 questions with collapsible Solution and Expert Solution. Tap a button to reveal the working.

IV. Long Answer Questions (Exercise 8.4)

Q 8.1

If cscθ+cotθ=p, then prove that cosθ=p²-1p²+1.

Q 8.2

Prove that √sec²θ+csc²θ=tanθ+cotθ.

Q 8.3

If 1+sin²θ=3sinθ, then prove that tanθ=1 or tanθ=12.

Q 8.4

Given that sinθ+2cosθ=1, then prove that 2sinθ-cosθ=2.

Q 8.5

If tanθ+secθ=l, then prove that secθ=l²+12l.

Q 8.6

If sinθ+cosθ=p and secθ+cscθ=q, then prove that q(p²-1)=2p.

Q 8.7

If asinθ+bcosθ=c, then prove that acosθ-bsinθ=√a²+b²-c².

Q 8.8

Prove that 1+secθ-tanθ1+secθ+tanθ=1-sinθcosθ.

Student Feedback

In a Collegedunia survey of 1,240 Class 10 students, 82% said the Trigonometry Exemplar needs a deeper grip on identities than the textbook exercises. 4 out of 5 who finished all four exercises felt more confident with trigonometry in CBSE board papers.

NCERT Exemplar Class 10 Maths Chapter 8 Introduction to Trigonometry FAQs

Ques. Where can I download the NCERT Exemplar Class 10 Maths Chapter 8 Solutions for free?

Ans. You can download the NCERT Exemplar Class 10 Maths Chapter 8 Introduction to Trigonometry Solutions PDF directly from this page using the red Download button. It is free and aligned to the 2026-27 CBSE syllabus.

Ques. How many problems are there in the Trigonometry Exemplar, and what types are they?

Ans. Chapter 8 has 55 Exemplar problems: 20 MCQs in Exercise 8.1, 10 true-or-false justification questions in Exercise 8.2, 17 short-answer problems in Exercise 8.3, and 8 long-answer identity proof questions in Exercise 8.4.

Ques. What are the main topics tested in the Trigonometry Exemplar?

Ans. The four main areas are: trigonometric ratios of acute angles in a right triangle, standard angle values (0°, 30°, 45°, 60°, 90°), complementary angle relations such as sin(90°-A) = cos A, and the three Pythagorean identities (sin²A + cos²A = 1; 1 + tan²A = sec²A; 1 + cot²A = cosec²A). Students should also know the reciprocal relations between sin/cosec, cos/sec, and tan/cot.

Ques. How is the Trigonometry Exemplar harder than the NCERT textbook for this chapter?

Ans. The textbook asks you to apply one technique at a time: evaluate a ratio, use one complementary relation, or verify a simple identity. The Exemplar needs multi-step reasoning. Exercise 8.2 wants written justifications with counterexamples, Exercise 8.3 combines two or more identities in one simplification, and Exercise 8.4 asks for full identity proofs where you choose the right Pythagorean identity at each step. The Exercise 8.1 MCQs also trap students who confuse complementary and supplementary angles.

Ques. What is the most common mistake students make in Chapter 8 Exemplar problems?

Ans. The most common mistake is treating tan 90° as a very large number instead of undefined. Because cos 90° = 0, the ratio sin 90°/cos 90° = 1/0 has no value in real numbers. A numerical value for tan 90° in a proof breaks the working. Always check whether the given angle makes a denominator zero first.

Ques. Are all three Pythagorean identities in the 2026-27 CBSE syllabus?

Ans. Yes. All three Pythagorean identities, sin²A + cos²A = 1, 1 + tan²A = sec²A, and 1 + cot²A = cosec²A, are part of the Class 10 Maths 2026-27 CBSE syllabus. The second and third identities are derived from the first by dividing through by cos²A and sin²A respectively. The Exemplar Exercise 8.4 proofs typically require students to choose which form to use at each step.

Ques. How much time should a Class 10 student spend on the Trigonometry Exemplar?

Ans. Plan about 3 to 4 hours total: roughly 40 minutes for the 20 MCQs, 35 minutes for the 10 true-or-false problems, 75 minutes for the 17 short-answer problems, and 60 minutes for the 8 long-answer proofs, plus a revision pass on anything you got wrong. If you know the standard angle values and the three Pythagorean identities by heart first, you will finish much faster.

NCERT Exemplar Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry

Watch Introduction to Trigonometry Class 10 Maths Explained

Exemplar Question Types at a Glance

Key Ratios, Identities & Angle Values

Standard Angle Values

Fundamental Trigonometric Identities

Complementary Angle Relations

How These Solutions Help You

Exemplar vs Textbook: Where It Gets Harder

Common Mistakes to Avoid

Other Resources for This Chapter

All Exemplar Questions with Step-by-Step Solutions

I. Multiple Choice Questions (Exercise 8.1)

Introduction to Trigonometry NCERT Exemplar

II. True / False with Reasoning (Exercise 8.2)

Introduction to Trigonometry NCERT Exemplar

III. Short Answer Questions (Exercise 8.3)

Introduction to Trigonometry NCERT Exemplar

IV. Long Answer Questions (Exercise 8.4)

Student Feedback

NCERT Exemplar Class 10 Maths Chapter 8 Introduction to Trigonometry FAQs

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