Exercise 8.2 Class 10 Maths NCERT Solutions — this page walks you through all four questions of the exercise, step by step. The whole exercise runs on Table 8.1, the standard-angle table built in Section 8.3: every value you substitute comes from that one table, for the angles 0°, 30°, 45°, 60° and 90°.
Question 1 evaluates five compound expressions, Question 2 picks options from four choices, Question 3 reads two tangent values back into angles, and Question 4 asks you to justify five true/false claims.
That makes the exercise fast — and easy to drop marks in. One swapped value, like writing sin 30° where you need sin 60°, sends an entire answer off. Every solution below shows the value it substitutes, the working that follows, and the specific slip to watch for, so you can check your own attempt against it.
Every question stem is reproduced word for word from the official NCERT Class 10 Mathematics textbook (rationalised edition) on ncert.nic.in, so you can compare any working with the source you are studying.
Exercise 8.2 Solutions
This exercise tests four skills in one sitting: direct evaluation of compound expressions (Q1), matching a simplified value against one of the six ratios (Q2, multiple choice), reversing a tangent value into an angle under a stated range (Q3), and justifying claims about how the ratios behave (Q4).
No identity beyond \(\sin^2 \theta + \cos^2 \theta = 1\) is needed anywhere — everything else is a read-off from Table 8.1, which is the page’s whole point: if you can read the table, you can solve the exercise.
Standard angle values from Table 8.1 you need here
Table 8.1 is the only source of values these four questions use. Keep it in front of you while you work — every answer below is a substitution from these six rows (NCERT, p. 125).
| ∠ A | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin A | 0 | \(\frac{1}{2}\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{\sqrt{3}}{2}\) | 1 |
| cos A | 1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{1}{2}\) | 0 |
| tan A | 0 | \(\frac{1}{\sqrt{3}}\) | 1 | \(\sqrt{3}\) | Not defined |
| cosec A | Not defined | 2 | \(\sqrt{2}\) | \(\frac{2}{\sqrt{3}}\) | 1 |
| sec A | 1 | \(\frac{2}{\sqrt{3}}\) | \(\sqrt{2}\) | 2 | Not defined |
| cot A | Not defined | \(\sqrt{3}\) | 1 | \(\frac{1}{\sqrt{3}}\) | 0 |
Two behaviour remarks carry the true/false items. As ∠A increases from 0° to 90°, sin A increases from 0 to 1, while cos A decreases from 1 to 0 (NCERT, p. 125). Question 4(ii) and 4(iii) are verdicts on exactly these two sentences — one is true, the other is false, and mixing them up is the classic slip.
Four cells of the table are not defined: tan 90°, sec 90°, cosec 0° and cot 0°. Each fails for the same reason: the ratio it needs as its denominator is zero there. For example, \(\cot 0^\circ = \cos 0^\circ/\sin 0^\circ = 1/0\) — division by zero. That reasoning decides Q4(v), and it is also why the tan 90° option appears as a distractor in Q2(ii).
Question 1: Evaluate the following :
(i) \( \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ \) (ii) \( 2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ \) (iii) \( \frac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ} \) (iv) \( \frac{\sin 30^\circ + \tan 45^\circ – \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ} \) (v) \( \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ – \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ} \)
Every term is a direct read from Table 8.1. Substitute the value first, then simplify the arithmetic — never simplify before you have written the substituted value down.
Part (i): sin 60° and cos 30° are a complementary pair — both equal \(\sqrt{3}/2\) — and sin 30° and cos 60° both equal \(1/2\).
\[ \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{1}{2}\cdot\frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1 \]
Part (ii): tan 45° = 1, so its square is 1.
Both cos²30° and sin²60° equal \((\sqrt{3}/2)^2 = 3/4\), so the last two terms cancel.
\[ 2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ = 2(1)^2 + \frac{3}{4} – \frac{3}{4} = 2 \]
Part (iii): cos 45° = \(1/\sqrt{2}\), sec 30° = \(2/\sqrt{3}\), and csc 30° = 2.
Combine the denominator, then rationalise the \(\sqrt{3}\) and the \(\sqrt{2}\).
\[ \frac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ} = \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2} = \frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}}} = \frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})} \]
\[ = \frac{\sqrt{3}(1-\sqrt{3})}{2\sqrt{2}(1-3)} = \frac{\sqrt{3}-3}{-4\sqrt{2}} = \frac{3-\sqrt{3}}{4\sqrt{2}} = \frac{3\sqrt{2}-\sqrt{6}}{8} \]
Part (iv): Substitute sin 30° = \(1/2\), tan 45° = 1, csc 60° = \(2/\sqrt{3}\), sec 30° = \(2/\sqrt{3}\), cos 60° = \(1/2\), cot 45° = 1, then rationalise the denominator by its conjugate.
\[ \frac{\frac{1}{2} + 1 – \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1} = \frac{\frac{3}{2} – \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{3}{2}} = \frac{3\sqrt{3} – 4}{4 + 3\sqrt{3}} \]
\[ = \frac{(3\sqrt{3}-4)(4-3\sqrt{3})}{(4+3\sqrt{3})(4-3\sqrt{3})} = \frac{24\sqrt{3} – 43}{16 – 27} = \frac{43 – 24\sqrt{3}}{11} \]
Part (v): cos²60° = \(1/4\), sec²30° = \(4/3\), tan²45° = 1, and the denominator \(\sin^2 30^\circ + \cos^2 30^\circ\) is exactly 1 by the identity \(\sin^2 \theta + \cos^2 \theta = 1\).
\[ \frac{5\cdot\frac{1}{4} + 4\cdot\frac{4}{3} – 1}{\frac{1}{4} + \frac{3}{4}} = \frac{\frac{5}{4} + \frac{16}{3} – 1}{1} = \frac{15 + 64 – 12}{12} = \frac{67}{12} \]
Final answers: (i) 1, (ii) 2, (iii) \(\dfrac{3\sqrt{2}-\sqrt{6}}{8}\), (iv) \(\dfrac{43-24\sqrt{3}}{11}\), (v) \(\dfrac{67}{12}\).
Students swap sin 30° (\(1/2\)) with sin 60° (\(\sqrt{3}/2\)) because the table rows look alike. Before substituting, point at the cell you are copying from — that single glance catches most slips. In part (iii), note that csc 60° and sec 30° happen to share the value \(2/\sqrt{3}\); if you wrote either one as \(\sqrt{3}/2\) you have it backwards.
Question 2: Choose the correct option and justify your choice :
(i) \( \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \)
- (A) \( \sin 60^\circ \)
- (B) \( \cos 60^\circ \)
- (C) \( \tan 60^\circ \)
- (D) \( \sin 30^\circ \)
(ii) \( \frac{1 – \tan^2 45^\circ}{1 + \tan^2 45^\circ} = \)
- (A) \( \tan 90^\circ \)
- (B) 1
- (C) \( \sin 45^\circ \)
- (D) 0
(iii) \( \sin 2A = 2 \sin A \) is true when \( A = \)
- (A) \( 0^\circ \)
- (B) \( 30^\circ \)
- (C) \( 45^\circ \)
- (D) \( 60^\circ \)
(iv) \( \frac{2 \tan 30^\circ}{1 – \tan^2 30^\circ} = \)
- (A) \( \cos 60^\circ \)
- (B) \( \sin 60^\circ \)
- (C) \( \tan 60^\circ \)
- (D) \( \sin 30^\circ \)
Each item collapses to a single standard value. Work the fraction down to its simplest form, then match the result against the six ratio columns of Table 8.1.
Part (i): \(\tan 30^\circ = 1/\sqrt{3}\), so its square is \(1/3\).
\[ \frac{2\tan 30^\circ}{1+\tan^2 30^\circ} = \frac{2/\sqrt{3}}{4/3} = \frac{2}{\sqrt{3}}\cdot\frac{3}{4} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \]
Since \(\sin 60^\circ = \sqrt{3}/2\), the answer is (A). The others fail: cos 60° = \(1/2\), tan 60° = \(\sqrt{3}\), sin 30° = \(1/2\) — none equal \(\sqrt{3}/2\).
Part (ii): \(\tan^2 45^\circ = 1\) because tan 45° = 1.
\[ \frac{1 – \tan^2 45^\circ}{1 + \tan^2 45^\circ} = \frac{1-1}{1+1} = \frac{0}{2} = 0 \]
The answer is (D). Option (A) is the trap: tan 90° is not defined, so it cannot equal this fraction — the value 0 belongs to the table’s defined cells only.
Part (iii): Test each option against \(\sin 2A = 2\sin A\).
- A = 0°: sin 0° = 0 and 2 sin 0° = 0 → equal ✓
- A = 30°: sin 60° = \(\sqrt{3}/2\) but 2 sin 30° = 1 → not equal ✗
- A = 45°: sin 90° = 1 but 2 sin 45° = \(\sqrt{2}\) → not equal ✗
- A = 60°: sin 120° = \(\sqrt{3}/2\) but 2 sin 60° = \(\sqrt{3}\) → not equal ✗
The answer is (A) 0°. The general double-angle identity \(\sin 2A = 2\sin A\cos A\) reduces to \(2\sin A\) only when cos A = 1, which the table gives only at A = 0°.
Part (iv): Observe the single difference from (i): the denominator now has a minus sign.
\[ \frac{2\tan 30^\circ}{1-\tan^2 30^\circ} = \frac{2/\sqrt{3}}{1-1/3} = \frac{2/\sqrt{3}}{2/3} = \frac{2}{\sqrt{3}}\cdot\frac{3}{2} = \frac{3}{\sqrt{3}} = \sqrt{3} \]
Since \(\tan 60^\circ = \sqrt{3}\), the answer is (C). This fraction matches the double-angle form \(2\tan\theta/(1-\tan^2\theta) = \tan 2\theta\), so \(\theta = 30^\circ\) gives \(\tan 60^\circ\).
In (i), students reach \((2/\sqrt{3})/(4/3)\) and mis-reduce: dividing by \(4/3\) means multiplying by \(3/4\), so write the inverted fraction before cancelling. In (iv), the only change from (i) is the sign in the denominator — \((1-1/3) = 2/3\) but \((1+1/3) = 4/3\). Read the sign before you substitute.
Question 3: If tan (A + B) = √3 and tan (A − B) = 1/√3 ; 0° < A + B ≤ 90° ; A > B , find A and B.
Read each tangent value backwards through Table 8.1 into an angle, apply the range condition to reject other candidates, then solve the two resulting equations together. This is the same pattern as Example 8 of the chapter, which used sin and cos of the angle-sums (NCERT, p. 127).
Step 1: \(\tan 60^\circ = \sqrt{3}\), and the range \(0^\circ < A+B \le 90^\circ\) leaves 60° as the only valid inverse.
So \(A + B = 60^\circ\).
Step 2: \(\tan 30^\circ = 1/\sqrt{3}\), and since A > B, the difference is positive.
So \(A – B = 30^\circ\).
Step 3: Add the two equations to eliminate B, subtract to eliminate A.
\[ (A+B) + (A-B) = 2A = 90^\circ \Rightarrow A = 45^\circ \]
\[ (A+B) – (A-B) = 2B = 30^\circ \Rightarrow B = 15^\circ \]
Final answer: A = 45° and B = 15°. Check: A = 45° > B = 15° ✓, and \(0^\circ < 60^\circ \le 90^\circ\) ✓.
The range condition is doing real work. Without \(0^\circ < A+B \le 90^\circ\), the equation \(\tan(A+B) = \sqrt{3}\) is ambiguous — \(\tan 240^\circ = \sqrt{3}\) too — but 240° violates the range, so 60° is forced. Students who drop the conditions and reach for 120° find \(\tan 120^\circ = -\sqrt{3}\), which does not even satisfy the given equation.
Write the conditions down before solving; they are part of the answer.
Question 4: State whether the following are true or false. Justify your answer.
(i) \( \sin (A + B) = \sin A + \sin B \).
(ii) The value of \( \sin \theta \) increases as \( \theta \) increases.
(iii) The value of \( \cos \theta \) increases as \( \theta \) increases.
(iv) \( \sin \theta = \cos \theta \) for all values of \( \theta \).
(v) \( \cot A \) is not defined for \( A = 0^\circ \).
Each statement is a claim about how the ratios behave, so the justification is either a counter-example (for a false claim) or the Table 8.1 remark (for a true one). A bare true/false never earns full marks — the question explicitly asks you to justify.
Part (i): False.
The correct identity for a sine of a sum is \(\sin(A+B) = \sin A \cos B + \cos A \sin B\), which is introduced later in the chapter — not \(\sin A + \sin B\).
A concrete counter-example makes it visible: take A = B = 30°.
Then \(\sin 60^\circ = \sqrt{3}/2 \approx 0.866\), but \(\sin 30^\circ + \sin 30^\circ = 1\).
The two sides differ, so the statement fails.
Part (ii): True.
From the remark under Table 8.1: as θ increases from 0° to 90°, sin θ increases from 0 to 1 (NCERT, p. 125).
The sine row reads 0, \(1/2\), \(1/\sqrt{2}\), \(\sqrt{3}/2\), 1 — strictly climbing the whole way.
Part (iii): False.
The same remark says cosine does the opposite: as θ goes from 0° to 90°, cos θ decreases from 1 to 0.
The cosine row reads 1, \(\sqrt{3}/2\), \(1/\sqrt{2}\), \(1/2\), 0 — falling, not rising.
For instance, \(\cos 30^\circ = \sqrt{3}/2\) but \(\cos 60^\circ = 1/2\), so a bigger angle gives a smaller cosine.
Part (iv): False.
\(\sin \theta = \cos \theta\) holds only at θ = 45°, where both sides equal \(1/\sqrt{2}\).
At any other acute angle they differ: \(\sin 30^\circ = 1/2\) while \(\cos 30^\circ = \sqrt{3}/2\).
So the statement is not true for all values of θ.
Part (v): True.
Since \(\cot A = \cos A/\sin A\), at A = 0° the denominator is \(\sin 0^\circ = 0\).
So \(\cot 0^\circ = 1/0\), which is division by zero — hence not defined.
This matches the not-defined cell in Table 8.1, and the chapter summary that cosec A and cot A are undefined wherever sin A = 0 (NCERT, p. 133).
Final verdicts: (i) False, (ii) True, (iii) False, (iv) False, (v) True.
Keep the two directions separate: sine rises while cosine falls. Students remember one and blindly copy it to the other — that exact swap is why both statements appear here. For the false items, a counter-example with actual values is the cleanest justification; for (ii) and (iii), quote the table row or the remark — that is the reasoning the examiner wants.
Method recap: checking your Exercise 8.2 Class 10 Maths NCERT Solutions
Four checking habits catch nearly every slip in this exercise:
- Write each standard value from Table 8.1 before substituting. A value written down can be checked against the table; a value carried in your head cannot.
- Handle reciprocals in pairs: csc A = \(1/\sin A\), sec A = \(1/\cos A\), cot A = \(1/\tan A\). Notice that csc 60° = \(2/\sqrt{3}\) and sec 30° = \(2/\sqrt{3}\) are two different cells that share a value — keep them distinct.
- Rationalise any denominator holding \(\sqrt{3}\): multiply top and bottom by the conjugate, as in Q1(iv) where the denominator \(4+3\sqrt{3}\) is cleared by \(4-3\sqrt{3}\).
- Use \(\sin^2 \theta + \cos^2 \theta = 1\) as a free check. Wherever the squared sine and cosine of the same angle appear together, they collapse to 1 — that is exactly Q1(v)’s denominator, and it saves a full line of arithmetic.
Micro-check with your own values: verify \(\cos 30^\circ + \sin 60^\circ = \sqrt{3}\) by substituting both terms as \(\sqrt{3}/2\). Each is the same number, so the sum is \(2\cdot\sqrt{3}/2 = \sqrt{3}\). This is not an exercise question — it just shows the value-discipline these problems demand.
Board papers keep reusing the evaluation and multiple-choice forms of Exercise 8.2, built on the 30°–45°–60° values of Table 8.1, so these substitution habits pay off in the exam directly.
The same standard values return in the next chapter, Some Applications of Trigonometry, where they solve heights-and-distances problems, and the slope idea you meet in the Coordinate Geometry chapter is literally the tangent of an angle. Exercise 8.3 continues the chapter by proving identities with these same values, so the habits you build here carry straight forward.
Exercise 8.2 is one exercise inside the wider Class 10 Maths notes and solutions collection, itself part of the Class 10 study material in the complete CBSE revision notes library.
FAQs: common doubts in Exercise 8.2
Why is cot 0° not defined in Question 4 (v)?
Because \(\cot A = \cos A/\sin A\), and at A = 0° the denominator is zero. \(\cos 0^\circ = 1\) and \(\sin 0^\circ = 0\), so \(\cot 0^\circ = 1/0\), which is division by zero and has no value. The same logic makes cosec 0° undefined, since \(\csc 0^\circ = 1/\sin 0^\circ\). Both cells are marked “not defined” in Table 8.1.
In Question 1 (iv), how do I rationalise the denominator containing √3?
Multiply the numerator and denominator by the conjugate of the denominator. The denominator is \(4+3\sqrt{3}\), so multiply top and bottom by \(4-3\sqrt{3}\). The product \((4+3\sqrt{3})(4-3\sqrt{3}) = 16 – 27 = -11\) is free of \(\sqrt{3}\), and the numerator becomes \(24\sqrt{3} – 43\). Dividing gives \((43 – 24\sqrt{3})/11\) — keep the minus sign in front.
Why is sin 2A = 2 sin A true only when A = 0°?
The general double-angle identity is \(\sin 2A = 2\sin A\cos A\). For \(\sin 2A\) to equal \(2\sin A\), the factor cos A must be 1. In the standard-angle table, cos A = 1 happens only at A = 0°. Checking all four options confirms it: at 30°, 45° and 60° the two sides differ; only 0° makes them equal.
Can I write sin 45° as √2/2 instead of 1/√2? Which answer is accepted in the board exam?
Yes — both are accepted. \(1/\sqrt{2}\) and \(\sqrt{2}/2\) are the same number: multiplying \(1/\sqrt{2}\) top and bottom by \(\sqrt{2}\) gives \(\sqrt{2}/2\). Board marking accepts either form, so you do not need to keep rationalising back and forth. Pick the form you can substitute fastest and stay consistent with it.
Why must A + B stay between 0° and 90° in Question 3?
Because the tangent function repeats: \(\tan 60^\circ = \sqrt{3}\), but \(\tan 240^\circ = \sqrt{3}\) as well. The range \(0^\circ < A+B \le 90^\circ\) identifies exactly one angle whose tangent is \(\sqrt{3}\), namely 60°, so the inverse is unique. The companion condition A > B fixes \(A – B = 30^\circ\) as positive rather than \(-30^\circ\) — together the two conditions pin down A and B completely.
Reference: NCERT Class 10 Mathematics textbook, chapter Introduction to Trigonometry.
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