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Mathematics · Introduction to Trigonometry · NCERT Exercises

Exercise 8.2

Complete, independently verified solutions for NCERT Exercise 8.2.

Mathematics · Chapter 8 · NCERT Exercises

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Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.

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Question

Question 1

special angles · evaluation

Evaluate: (i) sin60° cos30° + sin30° cos60°; (ii) 2tan²45° + cos²30° − sin²60°; (iii) [cos45° sin30° + tan45°]/[sec30° + cosec30°]; (iv) [sin30° + tan45° − cosec60°]/[sec30° + cos60° + cot45°]; (v) [5cos²60° + 4sec²30° − tan²45°]/[sin²30° + cos²30°].

Solution

Board-exam working:

(i)

32⋅32+12⋅12=34+14=1\frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2}+\frac12\cdot\frac12=\frac34+\frac14=1

(ii)

2(1)2+(32)2−(32)2=22(1)^2+\left(\frac{\sqrt3}{2}\right)^2-\left(\frac{\sqrt3}{2}\right)^2=2

(iii)

12⋅12+123+2=122+123+2\frac{\frac1{\sqrt2}\cdot\frac12+1}{\frac2{\sqrt3}+2}=\frac{\frac1{2\sqrt2}+1}{\frac2{\sqrt3}+2}

(iv)

12+1−2323+12+1=33−433+4\frac{\frac12+1-\frac2{\sqrt3}}{\frac2{\sqrt3}+\frac12+1}=\frac{3\sqrt3-4}{3\sqrt3+4}

(v)

5(12)2+4(23)2−1(12)2+(32)2\frac{5\left(\frac12\right)^2+4\left(\frac2{\sqrt3}\right)^2-1}{\left(\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}
=54+163−11=6712=\frac{\frac54+\frac{16}{3}-1}{1}=\frac{67}{12}

Final answer

(i) 1; (ii) 2;

(iii) 122+123+2;(iv) 33−433+4;(v) 6712\text{(iii) }\frac{\frac1{2\sqrt2}+1}{\frac2{\sqrt3}+2};\quad \text{(iv) }\frac{3\sqrt3-4}{3\sqrt3+4};\quad \text{(v) }\frac{67}{12}

Question

Question 2

multiple choice · special angles

Choose the correct option and justify: (i) 2tan30°/(1 + tan²30°); (ii) (1 − tan²45°)/(1 + tan²45°); (iii) sin 2A = 2 sin A is true when A is 0°, 30°, 45° or 60°; (iv) 2tan30°/(1 − tan²30°).

Solution

Board-exam working:

(i)

2⋅131+13=32=sin⁡60∘\frac{2\cdot\frac1{\sqrt3}}{1+\frac13}=\frac{\sqrt3}{2}=\sin60^\circ

Option A.

(ii)

1−11+1=0\frac{1-1}{1+1}=0

Option D.

(iii) Test A = 0°:

sin⁡(2⋅0∘)=0=2sin⁡0∘\sin(2\cdot0^\circ)=0=2\sin0^\circ

Option A.

(iv)

2⋅131−13=3=tan⁡60∘\frac{2\cdot\frac1{\sqrt3}}{1-\frac13}=\sqrt3=\tan60^\circ

Option C.

Final answer

(i) A; (ii) D; (iii) A; (iv) C.

Question

Question 3

tan values · simultaneous equations

If tan(A + B) = √3 and tan(A − B) = 1/√3, where 0° < A + B ≤ 90° and A > B, find A and B.

Solution

Board-exam working:

tan⁡(A+B)=3⇒A+B=60∘\tan(A+B)=\sqrt3\Rightarrow A+B=60^\circ
tan⁡(A−B)=13⇒A−B=30∘\tan(A-B)=\frac1{\sqrt3}\Rightarrow A-B=30^\circ

Adding:

2A=90∘⇒A=45∘2A=90^\circ\Rightarrow A=45^\circ

Subtracting:

2B=30∘⇒B=15∘2B=30^\circ\Rightarrow B=15^\circ

Final answer

A = 45° and B = 15°.

Question

Question 4

true or false · special angles

State true or false and justify: (i) sin(A + B) = sin A + sin B; (ii) sin θ increases as θ increases; (iii) cos θ increases as θ increases; (iv) sin θ = cos θ for all θ; (v) cot A is not defined for A = 0°.

Solution

Board-exam working:

(i) False. Take A = B = 30°.

sin⁡60∘=32≠1=sin⁡30∘+sin⁡30∘\sin60^\circ=\frac{\sqrt3}{2}\neq1=\sin30^\circ+\sin30^\circ

(ii) True for 0° ≤ θ ≤ 90°, as shown by the standard-value table.

(iii) False. cos θ decreases from 1 to 0 as θ increases from 0° to 90°.

(iv) False. Equality holds at 45°, but not for every angle.

(v) True.

cot⁡0∘=cos⁡0∘sin⁡0∘=10\cot0^\circ=\frac{\cos0^\circ}{\sin0^\circ}=\frac10

Division by zero is undefined.

Final answer

(i) False; (ii) True; (iii) False; (iv) False; (v) True.