# NCERT solution for class 10 maths introduction to trigonometry ( Chapter 8)

#### Solution for Exercise -8.1

Q1. In $$∆ \ ABC$$ , right angled at B, AB = 24 cm, BC = 7 cm. Determine :

(i) sin A, cos A
(ii) sin C, cos C

Given, in $$∆ \ ABC$$,
AB = 24 cm, BC = 7 cm and right angled at B

By using the Pythagoras theorem, we have

$$AC^2 \ = \ AB^2 \ + \ BC^2$$
=> $$AC^2 \ = \ (24)^2 \ + \ (7)^2$$
=> $$AC^2 \ = \ 576 \ + \ 49 \ = \ 625$$

=>   $$AC \ = \ \sqrt{625} \ = \ 25$$ cm

(i) $$sinA \ = \ \frac{BC}{AC} \ = \ \frac{7}{25}$$      [∵ $$sin\theta \ = \ \frac{P}{H}$$ ]

$$cosA \ = \ \frac{AB}{AC} \ = \ \frac{24}{25}$$      [∵ $$cos\theta \ = \ \frac{B}{H}$$ ]

(ii) $$sinC \ = \ \frac{AB}{AC} \ = \ \frac{24}{25}$$

$$cosC \ = \ \frac{BC}{AC} \ = \ \frac{7}{25}$$

Q2. In Fig. find tan P – cot R.

By using the Pythagoras theorem, we have

$$PR^2 \ = \ PQ^2 \ + \ QR^2$$

=> $$13^2 \ = \ 12^2 \ + \ QR^2$$

=> $$QR^2 \ = \ 13^2 \ - \ 12^2 \ = \ 169 \ - \ 144 \ = \ 25$$

=> $$QR \ = \ \sqrt{25} \ = \ 5$$

∴ $$tanP \ = \ \frac{P}{B} \ = \ \frac{QR}{PQ} \ = \ \frac{5}{12}$$
$$cotR \ = \ \frac{B}{P} \ = \ \frac{QR}{PQ} \ = \ \frac{5}{12}$$

Hence, $$tan P \ – \ cot R \ = \ \frac{5}{12} \ - \ \frac{5}{12} \ = \ 0$$

Q3. If $$sinA \ = \ \frac{3}{4}$$ , calculate cos A and tan A.

Given,
$$sinA \ = \ \frac{3}{4} \ = \ \frac{P}{H} \ = \ \frac{BC}{CA}$$

Let BC = 3k and AC = 4k.

Then, $$AB \ = \ \sqrt{AC^2 - BC^2} \ = \ \sqrt{(4k)^2 - (3k)^2} \ = \ \sqrt{16k^2 - 9k^2} \ = \ \sqrt{7k^2} \ = \ \sqrt{7}k$$

∴ $$cosA \ = \ \frac{B}{H} \ = \ \frac{AB}{AC} \ = \ \frac{ \sqrt{7}k}{4k} \ = \ \frac{ \sqrt{7}}{4}$$

$$tanA \ = \ \frac{P}{B} \ = \ \frac{BC}{AB} \ = \ \frac{3k}{ \sqrt{7}k} \ = \ \frac{3}{ \sqrt{7}}$$

Q4. Given 15 cot A = 8, find sin A and sec A.

Given,
$$15 cot A \ = \ 8$$
=> $$cotA \ = \ \frac{8}{15} \ = \ \frac{B}{P} \ = \ \frac{AB}{BC}$$

Let AB = 8k and BC = 15k.

Then, by Pythagoras theorem

$$AC \ = \ \sqrt{AB^2 + BC^2} \ = \ \sqrt{(8k)^2 + (15k)^2} \ = \ \sqrt{ 64k^2 + 225k^2} \ = \ \sqrt{289k^2} \ = \ 17k$$

∴ $$sinA \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{15k}{17k} \ = \ \frac{15}{17}$$

and, $$secA \ = \ \frac{H}{B} \ = \ \frac{AC}{AB} \ = \ \frac{17k}{8k} \ = \ \frac{17}{8}$$

Q5. Given $$sec\theta \ = \ \frac{13}{12}$$ , calculate all other trigonometric ratios.

Given, $$sec\theta \ = \ \frac{13}{12} \ = \ \frac{H}{B} \ = \ \frac{AC}{AB}$$

Let AC = 13k and AB = 12k.

Then by Pythagoras theorem,

$$BC \ = \ \sqrt{AC^2 - AB^2} \ = \ \sqrt{(13k)^2 - (12k)^2} \ = \ \sqrt{169k^2 - 144k^2} \ = \ \sqrt{25k^2} \ = \ 5k$$

∴ $$sin\theta \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{5k}{13k} \ = \ \frac{5}{13}$$

$$cos\theta \ = \ \frac{B}{H} \ = \ \frac{AB}{AC} \ = \ \frac{12k}{13k} \ = \ \frac{12}{13}$$

$$tan\theta \ = \ \frac{P}{B} \ = \ \frac{BC}{AB} \ = \ \frac{5}{12}$$

$$cot\theta \ = \ \frac{1}{tan\theta} \ = \ \frac{12}{5}$$

$$cosec\theta \ = \ \frac{1}{sin\theta} \ = \ \frac{13}{5}$$

Q6. If $$∠\ A$$ and $$∠\ B$$ are acute angles such that cos A = cos B, then show that $$∠ \ A \ = \ ∠ \ B$$.

In $$∆ \ ABC$$,
$$cosA \ = \ \frac{AC}{AB} \ = \ \frac{B}{H}$$

and, $$cosB \ = \ \frac{BC}{AB}$$

But given that, cos A = cos B

=> $$\frac{AC}{AB} \ = \ \frac{BC}{AB}$$
=> $$AC \ = \ BC$$

∴ $$∠ \ A \ = \ ∠ \ B$$ [∵ Angles opposite to equal sides are equal]

Q7 If $$cot\theta \ = \ \frac{7}{8}$$ , evaluate :

(i) $$\frac{(1+sin\theta)(1- sin\theta)}{(1+cos\theta)(1 - cos\theta)}$$

(ii) $$cot^2\theta$$

Given, $$cot\theta \ = \ \frac{B}{P} \ = \ \frac{AB}{BC} \ = \ \frac{7}{8}$$

Let AB = 7k and BC = 8k.
Then by Pythagoras theorem,

$$AC \ = \ \sqrt{BC^2 + AB^2} \ = \ \sqrt{(8k)^2 + (7k)^2} \ = \ \sqrt{64k^2 + 49k^2} \ = \ \sqrt{113k^2} \ = \ \sqrt{113}k$$

∴ $$sin\theta \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{8k}{ \sqrt{113}k} \ = \ \frac{8}{\sqrt{113}}$$

and, $$cos\theta \ = \ \frac{B}{H} \ = \ \frac{AB}{AC} \ = \ \frac{7k}{\sqrt{113}k} \ = \ \frac{7}{\sqrt{113}}$$

(i) ∴ $$\frac{(1+sin\theta)(1- sin\theta)}{(1+cos\theta)(1- cos\theta)} \ = \ \frac{1-sin^2\theta}{1 - cos^2\theta}$$

= $$\frac{1 - \frac{64}{113}}{1- \frac{49}{113}}$$

= $$\frac{113 - 64}{113 - 49}$$

= $$\frac{49}{64}$$

(ii) $$cot^2\theta \ = \ ( \frac{7}{8} )^2 \ = \ \frac{49}{64}$$

Q8. If $$3 cot A \ = \ 4$$, check whether $$\frac{1- tan^2A}{1+tan^2A} \ = \ cos^2A \ - \ sin^2A$$ or not.

Given, $$3 cot A \ = \ 4 \ = \ \frac{4}{3} \ = \ \frac{B}{P} \ = \ \frac{AB}{BC}$$

Let AB = 4k and BC = 3k

Then by Pythagoras theorem,

$$AC \ = \ \sqrt{AB^2 + BC^2} \ = \ \sqrt{(4k)^2 + (3k)^2} \ = \ \sqrt{16k^2 + 9k^2} \ = \ \sqrt{25k^2} \ = \ 5k$$

∴ $$sinA \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{3k}{5k} \ = \ \frac{3}{5}$$

$$cosA \ = \ \frac{B}{H} \ = \ \frac{AB}{AC} \ = \ \frac{4k}{5k} \ = \ \frac{4}{5}$$

and, $$tanA \ = \ \frac{1}{cotA} \ = \ \frac{3}{4}$$

Now, L.H.S. = $$\frac{1- tan^2A}{1+tan^2A} \ = \ \frac{1- \frac{9}{16}}{1+ \frac{9}{16}} \ = \ \frac{16 - 9}{16+9} \ = \ \frac{7}{25}$$

R.H.S. = $$cos^2A \ - \ sin^2A \ = \ ( \frac{4}{5})^2 \ - \ ( \frac{3}{5})^2 \ = \ \frac{16} {25} \ - \ \frac{9}{25} \ = \ \frac{7}{25}$$

∵ L.H.S = R.H.S.

∴ $$\frac{1 - tan^2A}{1+tan^2A} \ = \ cos^2A \ - \ sin^2A$$

Q9. In $$∆ \ ABC$$ right angled at B, if $$tan A \ = \ \frac{1}{ \sqrt{3}}$$ , find the value of

(i) $$sin A cos C \ + \ cos A sin C$$
(ii) $$cos A cos C - sin A sin C$$

Given, $$tanA \ = \ \frac{P}{H} \ = \ \frac{BC}{AB} \ = \ \frac{1}{\sqrt{3}}$$

Let $$BC \ = \ k$$ and $$AB \ = \ \sqrt{3} k$$.

Then by Pythagoras theorem,

$$AC \ = \ \sqrt{AB^2 + BC^2} \ = \ \sqrt{3k^2 + k^2} \ = \ \sqrt{4k^2} \ = \ 2k$$

∴ $$sinA \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{k}{2k} \ = \ \frac{1}{2}$$

$$cosA \ = \ \frac{B}{H} \ = \ \frac{AB}{AC} \ = \ \frac{ \sqrt{3}k}{2k} \ = \ \frac{ \sqrt{3}}{2}$$

$$sinC \ = \ \frac{P}{H} \ = \ \frac{AB}{AC} \ = \ \frac{ \sqrt{3}k}{2k} \ = \ \frac{ \sqrt{3}}{2}$$

and, $$cosC \ = \ \frac{B}{H} \ = \ \frac{BC}{AC} \ = \ \frac{k}{2k} \ = \ \frac{1}{2}$$

(i) $$sinA cosC \ + \ cosA sinC \ = \ \frac{1}{2} × \frac{1}{2} \ + \ \frac{ \sqrt{3}}{2} × \frac{ \sqrt{3}}{2} \ = \ \frac{1}{4} \ + \ \frac{3}{4} \ = \ \frac{4}{4} \ = \ 1$$

(ii) $$cos A cosC \ - \ Sin A sin C \ = \ \frac{ \sqrt{3}}{2} × \frac{1}{2} \ - \ \frac{1}{2} × \frac{ \sqrt{3}}{2} \ = \ 0$$

Q10. In $$∆ \ PQR$$ , right angled at Q, $$PR \ + \ QR \ = \ 25$$ cm and PQ = 5 cm. Determine the values of sin P, cos P , and tan P.

Given, PR + QR = 25 cm and PQ = 5cm

Let QR = x cm
Thus, PR = (25 – x) cm

Then by Pythagoras theorem,

$$RP^2 \ = \ RQ^2 \ + \ QP^2$$

$$(25-x)^2 \ = \ x^2 \ + \ 5^2$$

$$625\ - \ 50x \ + \ x^2 \ = \ x^2 \ + \ 25$$

$$- 50x \ = \ -600$$

=> $$x \ = \ \frac{600}{50} \ = \ 12$$

∴ RQ = 12 cm

=> RP = (25 – 12) = 13 cm

Now, $$sinP \ = \ \frac{P}{H} \ = \ \frac{RQ}{RP} \ = \ \frac{12}{13}$$

$$cosP \ = \ \frac{B}{H} \ = \ \frac{PQ}{RP} \ = \ \frac{5}{13}$$

and $$tanP \ = \ \frac{P}{B} \ = \ \frac{RQ}{PQ} \ = \ \frac{12}{5}$$

Q11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.

(ii) $$secA \ = \ \frac{12}{5}$$ for some value of angle A .

(iii) $$cos A$$ is the abbreviation used for the cosecant of angle A.

(iv) $$cot A$$ is the product of cot and A.

(v) $$sin\theta \ = \ \frac{4}{3}$$ for some angle $$\theta$$.

(i) False because sides of a right triangle may have any length, so tan A may have any value.

(ii) True as sec A is always greater than 1.

As, $$sec A \ = \ \frac{H}{B}$$ and Hypotenuse is the largest side ∴ sec A is always greater than 1

(iii) False as cos A is the abbreviation used for cosine A.

(iv) False as cot A is not the product of 'cot' and A. 'cot' separated from A has no meaning.

(v) False as sin $$\theta$$ cannot be > 1

#### Solution for Exercise -8.2

Q1. Evaluate :
(i) $$sin 60 ° cos 30° + sin 30° cos 60°$$

(ii) $$2tan^245° + cos^230° - sin^260°$$

(iii) $$\frac{cos45°}{sec30° + cosec30°}$$

(iv) $$\frac{sin30° + tan45°- cosec60°}{sec30° +cos60° +cot45°}$$

(v) $$\frac{5cos^260° +4sec^230° - tan^245°}{sin^230° + cos^230°}$$

(i) $$sin 60° cos 30° \ + \ sin 30° cos 60 \ = \ \frac{ \sqrt{3}}{2} × \frac{ \sqrt{3}}{2} \ + \ \frac{1}{2} × \frac{1}{2} \ = \ \frac{3}{4} \ + \ \frac{1}{4} \ = \ \frac{3+1}{4} \ = \ 1$$

(ii)$$2tan^245° \ + \ cos^230° \ - \ sin^260° \ = \ 2(1)^2 \ + \ ( \frac{ \sqrt{3}}{2})^2 \ - \ ( \frac{ \sqrt{3}}{2})^2 \ = \ 2 \ + \ \frac{3}{4} \ - \ \frac{3}{4} \ = \ 2$$

(iii) $$\frac{cos45°}{sec30° + cosec30°} \ = \ \frac{ \frac{1}{ \sqrt{2}}}{ \frac{2}{ \sqrt{3}} + 2} \ = \ \frac{ \frac{1}{ \sqrt{2}}}{ \frac{2 + 2 \sqrt{3}}{ \sqrt{3}}} \ = \ \frac{1}{ \sqrt{2}} × \frac{ \sqrt{3}}{2+2 \sqrt{3}} \ = \ \frac{ \sqrt{3}}{ \sqrt{2} × 2( \sqrt{3} + 1)} × \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1} \ = \ \frac{ \sqrt{3}( \sqrt{3} - 1)}{ \sqrt{2} × 2(3 - 1)} \ = \ \frac{ \sqrt{2} × \sqrt{3}( \sqrt{3} - 1)}{ \sqrt{2} × \sqrt{2} × 2 × 2} \ = \ \frac{3 \sqrt{2} - \sqrt{6}}{8}$$

(iv) $$\frac{sin30° + tan45°- cosec60°}{sec30° +cos60° +cot45°}$$ $$= \ \frac{ \frac{1}{2} + 1 - \frac{2}{ \sqrt{3}}}{ \frac{2}{ \sqrt{3}} + \frac{1}{2} + 1}$$ $$= \ \frac{ \frac{1 + 2}{2} - \frac{2}{ \sqrt{3}}}{ \frac{2}{ \sqrt{3}} + \frac{1 + 2}{2}}$$ $$= \ \frac{ \frac{3}{2} - \frac{2}{ \sqrt{3}}}{ \frac{2}{ \sqrt{3}} + \frac{3}{2}}$$ $$= \ \frac{ \frac{3 \sqrt{3} - 4}{2 \sqrt{3}}}{ \frac{4 + 3 \sqrt{3}}{2 \sqrt{3}}}$$ $$= \ \frac{3 \sqrt{3} - 4}{4 +3 \sqrt{3}}$$ $$= \ \frac{(3 \sqrt{3} - 4)(3 \sqrt{3} - 4)}{(4 + 3 \sqrt{3})(3 \sqrt{3} - 4)}$$ $$= \ \frac{27 + 16 - 12 \sqrt{3} - 12 \sqrt{3}}{27- 16}$$ $$= \ \frac{43 - 24 \sqrt{3}}{11}$$

(v) $$\frac{5cos^260° +4sec^230° - tan^245°}{sin^230° + cos^230°}$$ $$= \ \frac{5( \frac{1}{2})^2 + 4( \frac{2}{ \sqrt{3}})^2 - (1)^2}{( \frac{1}{2})^2 + ( \frac{ \sqrt{3}}{2})^2}$$ $$= \ \frac{5 × \frac{1}{4} + 4 × \frac{4}{3} - 1}{ \frac{1}{4} + \frac{3}{4}}$$ $$= \ \frac{5}{4} \ + \ \frac{16}{3} - 1$$ $$= \ \frac{15 + 64 - 12}{12} \ = \ \frac{67}{12}$$

Q2. Choose the correct option and justify :

(i) $$\frac{2tan30°}{1+tan^230°}$$ =

(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°

(ii) $$\frac{1 - tan^245 °}{1+tan^245 °}$$ =

(A) tan 90°
(B) 1
(C) sin 45 °
(D) 0

(iii) $$sin 2A \ = \ 2 sinA$$ is true when A =

(A) 0°
(B) 30°
(C) 45°
(D) 60°

(iv) $$\frac{2tan^230 °}{1- tan^230 °}$$ =

(A) cos 60°
(B) sin 60°
(C) tan 60 °
(D) none of these

(i) (A), as
$$\frac{2tan30°}{1+tan^230°}$$ $$= \ \frac{2 × \frac{1}{ \sqrt{3}}}{1+( \frac{1}{ \sqrt{3}})^2} \ = \ \frac{ \frac{2}{ \sqrt{3}}}{1 + \frac{1}{3}}$$ $$= \ \frac{2}{ \sqrt{3}} × \frac{3}{4}$$ $$= \ \frac{ \sqrt{3}}{2} \ = \ sin60 °$$

(ii) (D), as
$$\frac{1- tan^245 °}{1+tan^245 °}$$ $$=\ \frac{1- 1}{1+1} \ = \ 0$$

(iii) (A), as
when A = 0, sin 2 A = sin 0 = 0

and, 2 sinA = 2 sin 0 = 2 × 0 = 0
=> sin 2A = 2sinA, when A = 0
(iv) (C), as
$$\frac{2tan^230 °}{1- tan^230 °}$$ $$= \ \frac{2 × \frac{1}{ \sqrt{3}}}{1 - ( \frac{1}{ \sqrt{3}})^2}$$ $$= \ \frac{ \frac{2}{ \sqrt{3}}}{1 - \frac{1}{3}}$$ $$= \ \frac{2}{ \sqrt{3}} × \frac{3}{3 - 1} \ = \ \frac{2}{ \sqrt{3}} × \frac{3}{2}$$ $$= \ \sqrt{3} \ = \ tan60°$$

Q3. If $$tan (A + B) \ = \ \sqrt{3}$$ and $$tan (A – B) \ = \ \frac{1}{ \sqrt{3}}$$
$$0° < \ A+B \ \le \ 90° \ ; \ A \ > \ B$$ , find A and B.

$$tan (A + B) \ = \ \sqrt{3}$$
=> $$\ tan(A+B) \ = \ tan60°$$
$$\ A+B \ = \ 60°$$ ............(1)

$$tan (A – B) \ = \ \frac{1}{ \sqrt{3}}$$
=> $$\ tan(A-B) \ = \ tan30°$$
$$A – B \ = \ 30°$$...............(2)

Solving (1) and (2), we get

$$\angle$$A = 45° and $$\angle$$B = 15°

Q4. State whether the following are true of 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°.

(i) False, because
for a case when A = 60° and B = 30°,

$$sin (A + B) \ = \ sin (60° + 30°) \ = \ sin 90° \ = \ 1$$

and, $$sin A \ + \ sin B \ = \ sin 60° \ + \ sin 30° \ = \frac{ \sqrt{3}}{2} \ + \ \frac{1}{2} \ = \ \frac{ \sqrt{3} + 1}{2}$$

$$sin (A + B) \ \ne \ sin A + sin B$$

(ii) True, as we can see from the table as the value of $$sin \theta$$ increases $$\theta$$ also increases.

(iii) False, as we can see from the table as the value of $$cos\theta$$ increases $$\theta$$ decreases.

(iv) False, as it is only true for $$\theta \ = \ 45°$$.

(v) True, as $$tan 0° \ = \ 0$$ and $$cot0° \ = \ \frac{1}{tan0°} \ = \ \frac{1}{0}$$ ,i.e., not defined.

#### Solution for Exercise -8.3

Q1. Evaluate :
(i) $$\frac{sin18°}{cos72°}$$

(ii) $$\frac{tan26°}{cot64°}$$

(iii) $$cos 48° – sin 42°$$

(iv) $$cosec 31° – sec 59°$$

(i) $$\frac{sin18°}{cos72°} \ = \ \frac{sin(90°- 72)}{cos72°}$$ $$= \ \frac{cos72°}{cos72°} \ = \ 1$$    [∵ $$sin(90° - \theta) \ = \ cos\theta$$ ]

(ii) $$\frac{tan26°}{cot64°} \ = \ \frac{tan(90° - 64°)}{cot64°}$$ $$= \ \frac{cot64°}{cot64°} \ = \ 1$$    [∵ $$tan(90° - \theta) \ = \ cot\theta$$ ]

(iii) $$cos48° – sin42° \ = \ cos (90° – 42°) – sin 42°$$ $$= \ sin 42° – sin 42° = 0$$    [∵ $$cos(90° - \theta) \ = \ sin\theta$$ ]

(iv) $$cosec 31° – sec 59° \ = \ cosec (90° – 59°) – sec 59°$$ $$= \ sec59° - sec59° \ = \ 0$$    [∵ $$cosec(90° - \theta) \ = \ sec\theta$$ ]

Q2. Show that
(i) $$tan 48° tan 23° tan 42° tan 67° \ = \ 1$$

(ii) $$cos 38° cos 52° \ - \ sin 38° sin 52 ° \ = \ 0$$

(i) $$tan 48° tan 23° tan 42 ° tan 67° \ = \ tan (90° – 42°) tan(90° – 67°) tan 42° tan 67°\ = \ cot 42° cot 67° tan 42° tan 67°$$    [∵ $$tan(90° - \theta) \ = \ cot\theta$$ ]

$$= \ \frac{1}{tan42°} × \frac{1}{tan67°} × tan42°tan67° \ = \ 1$$

(ii) $$cos 38°cos 52°– sin 38°sin 52° \ = \ cos(90° – 52°)cos 52° – sin(90°– 52°)sin 52°$$

$$= \ sin 52° cos 52° - cos 52° sin 52° \ = \ 0$$    [∵ $$cos(90°-\theta) \ = \ sin\theta$$ and $$sin(90°-\theta) \ = \ cos\theta$$ ]

Q3. If $$tan 2A \ = \ cot (A – 18°)$$, where 2A is an acute angle, find the value of A.

Given,

$$tan 2A = cot (A – 18°)$$

=>$$\ cot(90°–2A) = cot (90° – 18)$$    [ ∵ $$tan \theta \ = \ cot (90° – \theta )$$ ]

=> $$90° - 2A \ = \ A – 18°$$    [ ∵ (90° – 2A) and (A – 18°) are both acute angle ]

=> $$2A + A \ = \ 18° + 90°$$

=> $$3A \ = \ 108°$$

=> $$A \ = \ 36°$$

Q4. If $$tan A \ = \ cot B$$ , prove that $$A + B \ = \ 90°$$.

Given, $$tan A \ = \ cot B$$
We know that, $$cot \theta \ = \ tan(90° – \theta)$$

=> $$cot B \ = \ tan(90° - B)$$

∴ $$tan A \ = \ tan(90° - B)$$

=> $$(90° – A) \ = \ B$$    [∵ (90° – A) and B are both acute angles ]

=> $$A + B \ = \ 90°$$

Q5. If $$sec 4 A \ = \ cosec (A – 20°)$$ , where 4 A is an acute angle, find the value of A.

Given, $$sec 4A \ = \ cosec (A – 20°)$$
We know that, $$cosec(90° - \theta ) \ = \ sec\theta$$

=> $$cosec (90°– 4A) \ = \ sec 4A$$

∴ $$cosec (90° – 4A) \ = \ cosec (A – 20°)$$

=> $$90° – 4A \ = \ A – 20°$$    [∵ (90° – 4A) and (A – 20°) are both acute angles]

$$4A + A \ = \ 20° + 90°$$

=> $$5A \ = \ 110°$$

=> $$A \ = \ 22°$$

Q6. If A, B and C are interior angles of a $$∆ \ ABC$$, then show that
$$sin( \frac{B+C}{2}) \ = \ cos \frac{A}{2}$$

Given, A, B and C are the interior angles of a $$∆ \ ABC$$

We know that, sum of interior angles of a triangle is 180°

∴ $$A + B + C \ = \ 180°$$

=> $$\frac{A+B+C}{2} \ = \ 90°$$

=> $$\ \frac{B+C}{2} \ = \ 90° - \frac{A}{2}$$

=> $$sin( \frac{B+C}{2} ) \ = \ sin( 90° - \frac{A}{2} )$$

=> $$sin( \frac{B+C}{2} ) \ = \ cos \frac{A}{2}$$    [ ∵ $$sin(90° - \theta) \ = \ cos\theta$$ ]

Q7. Express $$sin 67° + cos 75°$$ in terms of trigonometric ratios of angles between 0° and 45°.

Given,

$$sin 67° + cos 75°$$

= $$sin (90° – 23) + cos (90° – 15°)$$

$$= \ cos 23° + sin 15°$$    [ ∵ $$sin( 90° - \theta) \ = \ cos \theta$$ and $$cos( 90° - \theta) \ = \ sin \theta$$

#### Solution for Exercise -8.4

Q.1 Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Consider a $$∆ \ ABC$$

$$cotA \ = \ \frac{B}{P} \ = \ \frac{AB}{BC} \ => \ \frac{AB}{BC} \ = \ \frac{cotA}{1}$$

Let $$AB = kcotA$$ and $$BC = k$$.

By Pythagoras Theorem,

$$AC \ = \ \sqrt{AB^2 + BC^2} \ = \ \sqrt{k^2cot^2A + k^2} \ = \ k \sqrt{1 + cot^2A}$$

∴ $$sinA \ = \ \frac{P}{B} \ = \ \frac{BC}{AC} \ = \ \frac{k}{k \sqrt{1+cot^2A}} \ = \ \frac{1}{ \sqrt{1+cot^2A}}$$

$$secA \ = \ \frac{H}{B} \ = \ \frac{AC}{AB} \ = \ \frac{k \sqrt{1+cot^2A}}{kcotA} \ = \ \frac{ \sqrt{1+cot^2A}}{cotA}$$

and, $$tanA \ = \ \frac{P}{B} \ = \ \frac{BC}{AB} \ = \ \frac{k}{kcotA} \ = \ \frac{1}{cotA}$$

Q2. Write the other trigonometric ratios of A in terms of secA.

We know that , $$cos A \ = \ \frac{1}{sec A}$$

Now, $$sin^2A + cos^2A \ = \ 1$$

=> $$sin^2A \ = \ 1 - cos^2A$$

=> $$sin A \ = \ \frac{ \sqrt{ sec^2A - 1}}{sec A}$$

$$tan^2A + 1 \ = \ sec^2A$$

=> $$tan A \ = \ \sqrt{sec^2A - 1}$$

$$cot A \ = \ \frac{1}{tan A}$$

=> $$cot A \ = \ \frac{1}{ \sqrt{sec^2A - 1}}$$

$$cosec A \ = \ \frac{1}{sin A}$$

=> $$cosec A \ = \ \frac{sec A}{ \sqrt{sec^2A - 1}}$$

Q3. Evaluate :

(i) $$\frac{sin^263° + sin^227°}{cos^217° + cos^273°}$$

(ii) $$sin 25° cos 65° \ + \ cos25° sin65°$$

(i) $$\frac{sin^263° + sin^227°}{cos^217° + cos^273°} \ = \ \frac{cos^227°+sin^227°}{ sin^273° + cos^273°}$$

[ ∵ $$sin(90° - \theta) \ = \ cos\theta$$ => $$sin63° \ = \ sin(90° – 27°) \ = \ cos27°$$ ]

and [∵ $$cos(90° - \theta) \ = \ sin\theta$$] => $$cos17° \ = \ cos(90° – 73°) \ = \ sin 73°$$ ]

$$= \ \frac{1}{1} \ = \ 1$$    [∵ $$cos^2A + sin^2A \ = \ 1$$ ]

(ii) $$sin25° cos65° + cos25°sin65°$$

= $$sin(90°– 65°). cos65° + cos(90°– 65°) sin65°$$

= $$cos65° cos65° + sin 65° sin65°$$    [∵ $$sin(90°-\theta ) \ = \ cos\theta$$ and $$cos(90°-\theta ) \ = \ sin\theta$$ ]

= $$cos^265° + sin^265° \ = \ 1$$

Q4. Choose the correct option. Justify your choice :

(i) $$9sec^2A - 9tan^2A \ =$$

(A) 1       (B) 9
(C) 8       (D) 0

(ii) $$(1+ tan\theta + sec\theta )(1 + cos \theta – cosec \theta) \ =$$

(A) 0       (B) 1
(C) 2       (D) None of these

(iii) $$(secA + tanA)(1 – sinA) \ =$$

(A) secA       (B) sinA
(C) cosecA       (D) cosA

(iv) $$\frac{1+tan^2A}{1+cot^2A} \ =$$

(A) $$sec^2A$$       (B) –1
(C) $$cot^2A$$       (D) none of these

(i) (B), as

$$9sec^2A - 9tan^2A$$
= $$9(sec^2A - tan^2A)$$
= $$9 × 1 \ = \ 9$$    [∵ $$1+tan^2A \ = \ sec^2A$$ ]

(ii) (C), as

$$(1+ tan\theta + sec\theta )(1 + cos \theta – cosec \theta)$$

$$= \ (1 \ + \ \frac{sin\theta}{cos\theta} \ + \ \frac{1}{cos\theta} )(1 \ + \ \frac{cos\theta}{sin\theta} \ - \ \frac{1}{sin\theta})$$

$$= \ ( \frac{cos\theta \ + \ sin\theta \ + \ 1}{cos\theta})( \frac{sin\theta \ + \ cos\theta \ - \ 1}{sin\theta})$$

$$= \ \frac{(cos\theta + sin\theta )^2 \ - \ 1}{sin\theta cos\theta}$$    [∵ $$(A+B)(A-B) \ = \ A^2 \ - \ B^2$$ ]

$$= \ \frac{cos^2 \theta \ + \ sin^2 \theta \ + \ 2cos\theta sin\theta \ - \ 1}{sin\theta cos\theta}$$    [∵ $$sin^2 \theta \ + \ cos^2 \theta \ = \ 1$$ ]

$$= \ \frac{1 \ + \ 2cos \theta sin \theta \ - \ 1}{sin \theta cos \theta} \ = \ \frac{2cos \theta sin\theta}{sin \theta cos \theta} \ = \ 2$$

(iii) (D), as

$$(secA + tanA) (1 – sinA) \ = \ ( \frac{1}{cosA} \ + \ \frac{sinA}{cosA})(1-sinA)$$

$$= \ ( \frac{1+sinA}{cosA})(1-sinA)$$    [ ∵ $$(A+B)(A-B) \ = \ A^2 \ - \ B^2$$ ]

$$= \ \frac{1 \ - \ sin^2A}{cosA} \ = \ \frac{cos^2A}{cosA} \ = \ cosA$$    [∵ $$sin^2A \ + \ cos^2A \ = \ 1$$ ]

(iv) (D), as

$$\frac{1+tan^2A}{1+cot^2A}$$ $$= \ \frac{sec^2A}{cosec^2A}$$    [ ∵ $$1+tan^2A \ = \ sec^2A$$ and $$1+cot^2A \ = \ cosec^2A$$ ]

$$= \ \frac{ \frac{1}{cos^2A}}{ \frac{1}{sinn^2A}}$$ $$= \ \frac{sin^2A}{cos^2A} \ = \ tan^2A$$

Q5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined :

(i) $$(cosec \theta - cot \theta )^2 \ = \ \frac{1 - cos \theta}{1 + cos \theta}$$

(ii) $$\frac{cosA}{1+sinA} \ + \ \frac{1+sinA}{cosA} \ = \ 2 secA$$

(iii) $$\frac{tan \theta}{1 - cot\theta} \ + \ \frac{cot \theta}{1 - tan\theta} \ = \ 1 \ + \ sec \theta cosec \theta$$

[Hint : Write the expression in terms of sin $$\theta$$ and cos $$\theta$$]

(iv) $$\frac{1 \ + \ secA}{secA} \ = \ \frac{sin^2A}{1 \ - \ cosA}$$

[Hint : Simplify LHS and RHS separately]

(v) $$\frac{cosA \ - \ sinA \ + \ 1}{ cosA \ + \ sinA \ - \ 1} \ = \ cosecA \ + \ cotA$$ , using the identity $$cosec^2A \ = \ 1 \ + \ cot^2A$$

(vi) $$\sqrt{ \frac{1 \ + \ sinA}{1 \ - \ sinA}} \ = \ secA \ + \ tanA$$

(vii) $$\frac{sin\theta - 2sin^3 \theta}{2cos^3 \theta - cos\theta} \ = \ tan \theta$$

(viii) $$(sinA \ + \ cosecA)^2 \ + \ (cosA \ + \ secA)^2 \ = \ 7 \ + \ tan^2A \ + \ cot^2A$$

(ix) $$(cosecA \ - \ sinA)(secA \ - \ cosA) \ = \ \frac{1}{tanA \ + \ cotA}$$

[Hint : Simplify LHS and RHS separately]

(x) $$( \frac{1 \ + \ tan^2A}{1 \ + \ cot^2A} ) \ = \ ( \frac{1 \ - \ tanA}{1 \ - \ cotA})^2 \ = \ tan^2A$$

(i) L.H.S. = $$(cosec \theta - cot \theta )^2 \ = \ ( \frac{1}{sin\theta} \ - \ \frac{cos \theta}{ sin \theta} )^2$$ $$= \ ( \frac{1 \ - \ cos \theta }{ sin \theta } )^2$$

$$= \ \frac{ ( 1 \ - \ cos \theta )^2 }{ sin^2 \theta} \ = \ \frac{(1 \ - \ cos \theta)^2 }{ 1 \ - \ cos^2 \theta}$$    [∵ $$sin^2 \theta \ = \ 1 \ - \ cos^2 \theta$$ ]

$$= \ \frac{(1 \ - \ cos\theta )^2}{(1 \ - \ cos\theta)(1 \ + \ cos\theta)} \ = \ \frac{1 \ - \ cos\theta}{1 \ + \ cos\theta}$$ = R.H.S. [∵ $$A^2 \ - \ B^2 \ = \ (A+B)(A-B)$$ ]

(ii) L.H.S. = $$\frac{cosA}{1+sinA} \ + \ \frac{1+sinA}{cosA}$$ $$= \ \frac{cos^2A \ + \ (1 \ + \ sinA)^2}{cosA(1 \ + \ sinA)}$$

$$= \ \frac{cos^2A \ + \ 1 \ + \ 2sinA \ + \ sin^2A}{cosA(1 \ + \ sinA)}$$ $$= \ \frac{(cos^2A \ + \ sin^2A) \ + \ 1 \ + \ 2sinA}{cosA(1 \ + \ sinA)}$$

$$= \ \frac{1 \ + \ 1 \ + \ 2sinA}{cosA(1 \ + \ sinA)}$$    [∵ $$sin^2A \ + \ cos^2A \ = \ 1$$ ]

$$= \ \frac{2 \ + \ 2sinA}{cosA(1 \ + \ sinA)} \ = \ \frac{2(1 \ + \ sinA)}{cosA(1 \ + \ sinA)}$$

$$= \ \frac{2}{cosA} \ = \ 2secA \ =$$ R.H.S.

(iii) L.H.S. = $$\frac{tan \theta}{1 - cot\theta} \ + \ \frac{cot \theta}{1 - tan\theta}$$ $$= \frac{ \frac{sin \theta}{cos\theta}}{ 1 \ - \ \frac{cos\theta}{sin\theta}} \ + \ \frac{ \frac{cos\theta}{sin\theta}}{1 \ - \ \frac{sin\theta}{cos\theta}}$$

$$\frac{sin^2 \theta}{cos\theta(sin\theta \ - \ cos\theta)} \ + \ \frac{cos^2 \theta}{sin\theta(cos\theta \ - \ sin\theta)}$$ $$= \ ( \frac{1}{sin\theta \ - \ cos\theta}) [ \frac{sin^2 \theta}{cos\theta} \ - \ \frac{cos^2 \theta}{sin\theta} ]$$

$$= \ \frac{sin^3 \theta \ - \ cos^3 \theta}{ cos\theta sin\theta( sin\theta \ - \ cos\theta)}$$ $$= \ \frac{(sin\theta \ - \ cos\theta)( sin^2 \theta \ + \ sin\theta cos\theta \ + \ cos^2 \theta)}{( cos\theta sin\theta( sin\theta \ - \ cos\theta)}$$    [ ∵ $$a^3 \ - \ b^3 \ = \ (a-b)(a^2 \ + \ ab \ + \ b^2)$$ ]

$$\frac{1 \ + \ sin\theta cos\theta}{sin\theta cos\theta}$$    [ ∵ $$sin^2 \theta \ + \ cos^2 \theta \ = \ 1$$ ]

$$= \ 1 \ + \ sec \theta cosec \theta$$ = R.H.S.

(iv) L.H.S. $$= \ \frac{1 \ + \ secA}{secA} \ = \ 1 \ + \ \frac{1}{secA} \ = \ 1 \ + \ cosA$$ $$= \ \frac{(1 \ - \ cosA )(1 \ + \ cosA)}{1 \ - \ cosA}$$

$$= \ \frac{1 \ - \ cos^2A}{1 \ - \ cosA}$$    [ ∵ $$A^2 \ - \ B^2 \ = \ (A+B)(A-B)$$ ]

$$\frac{sin^2A}{1 \ - \ cosA}$$    [ ∵ $$sin^2A \ = \ 1 \ - \ cos^2A$$ ]

= R.H.S.

(v) L.H.S. = $$\frac{cosA \ - \ sinA \ + \ 1}{ cosA \ + \ sinA \ - \ 1}$$ $$= \ \frac{ \frac{ cosA \ - \ sinA \ + \ 1}{sinA}}{ \frac{cosA \ + \ sinA \ - \ 1}{sinA}}$$

$$= \ \frac{cotA \ - \ 1 \ + \ cosecA}{cosA \ + \ 1 \ - \ cosecA}$$    [ ∵ $$1 \ + \ cot^2A \ = \ cosec^2A$$ ]

$$= \ \frac{cotA \ + \ cosecA \ - \ (cosec^2A \ - \ cot^2A)}{cotA \ - \ cosecA \ + \ 1}$$ $$= \ \frac{cotA \ + \ cosecA \ - \ (cosecA \ + \ cotA)(cosecA \ - \ cotA)}{cotA \ - \ cosecA \ + \ 1}$$ [ ∵ $$A^2 \ - \ B^2 \ = \ (A+B)(A-B)$$ ]

Taking common(cosecA + cotA)

$$= \ \frac{(cosecA \ + \ cotA)(1 \ - \ cosecA \ + \ cot)}{(cotA \ - \ cosecA \ + \ 1 )}$$ $$= \ cosec A \ + \ cot A$$ = R.H.S.

(vi) L.H.S. = $$\sqrt{ \frac{1 \ + \ sinA}{1 \ - \ sinA}}$$ $$= \ \sqrt{ \frac{1+sinA}{1-sinA} × \frac{1+sinA}{1+sinA}}$$    [Multiplying and dividing by $$\sqrt{1+sinA}$$ ]

$$= \ \sqrt{ \frac{(1+sinA)^2}{1-sin^2A}} \ = \ \sqrt{ \frac{(1+sinA)^2}{cos^2A}}$$ [∵ $$sin^2A \ + \ cos^2A \ = \ 1$$ ]

$$= \ \frac{1+sinA}{cosA}$$ $$= \ \frac{1}{cosA} \ + \ \frac{sinA}{cosA}$$

$$= \ secA \ + \ tanA$$ [∵ $$tanA \ = \ \frac{sinA}{cosA}$$ and $$secA \ = \ \frac{1}{cosA}$$ ]

= R.H.S.

(vii) L.H.S. = $$\frac{sin\theta - 2sin^3 \theta}{2cos^3 \theta - cos\theta}$$ $$= \ \frac{ sin \theta(1-2sin^2 \theta)}{cos \theta (2cos^2 \theta -1)}$$

$$= \ tan \theta [ \frac{1-sin^2 \theta - sin^2 \theta}{ cos^2 \theta + cos^2 \theta - 1} ]$$ $$= \ tan \theta [ \frac{ cos^2 \theta - sin^2 \theta }{ cos^2 \theta - sin^2 \theta } ]$$    [ ∵ $$sin^2 \theta \ + \ cos^2 \theta \ = \ 1$$ ]

$$= \ tan\theta$$

= R.H.S.

(viii) L.H.S. =$$(sinA \ + \ cosecA)^2 \ + \ (cosA \ + \ secA)^2$$ $$= \ (sin^2A \ + \ cosec^2A \ + \ 2sinAcosecA) \ + \ (cos^2A \ + \ sec^2A \ + \ 2cosAsecA)$$

$$= \ (sin^2A \ + \ cosec^2A \ + \ 2 ) \ + \ (cos^2A \ + \ sec^2A \ + \ 2 )$$ $$= \ 1+ cosec^2A \ + \ sec^2A \ + \ 4$$    [∵ $$sin^2A \ + \ cos^2A \ = \ 1$$ ]

$$= \ 1 \ + \ cot^2A \ + \ 1 \ + \ tan^2A \ + \ 5$$    [∵ $$sec^2A \ = \ 1 \ + \ tan^2A$$ and $$cosec^2A \ = \ 1 \ + \ cot^2A$$ ]

$$= \ 7 \ + \ tan^2A \ + \ cot^2A$$ = R.H.S.

(ix) L.H.S. = $$(cosecA \ - \ sinA)(secA \ - \ cosA)$$ $$= ( \frac{1}{sinA} \ - \ sinA)( \frac{1}{cosA} \ - \ cosA)$$

$$= \ ( \frac{1 \ - \ sin^2A}{sinA} )( \frac{1 \ - \ cos^2A}{cosA} )$$ $$= \frac{ cos^2A}{sinA} × \frac{sin^2A}{cosA}$$

$$= \ sinA cosA$$ $$= \ \frac{sinAcosA}{sin^2A \ + \ cos^2A}$$ [∵ $$sin^2A \ + \ cos^2A \ = \ 1$$ ]

Dividing Numerator and Denominator by $$sinA cosA$$, we get

$$\frac{ \frac{sinAcosA}{sinAcosA}}{ \frac{sin^2A}{sinAcosA} \ + \ \frac{cos^2A}{sinAcosA}}$$ $$= \ \frac{1}{ \frac{sinA}{cosA} \ + \ \frac{cosA}{sinA}}$$

$$= \ \frac{1}{tanA \ + \ cotA}$$ = R.H.S

(x) L.H.S. = $$( \frac{1 \ + \ tan^2A}{1 \ + \ cot^2A} )$$ $$= \ \frac{sec^2A}{cosec^2A}$$

$$= \ \frac{1}{cos^2A} \ × \ sin^2A \ = \ tan^2A$$    [ ∵ $$secA \ = \ \frac{1}{cosA}$$ and $$cosecA \ = \ \frac{1}{sinA}$$ ]

R.H.S. = $$\frac{1 \ - \ tanA}{1 \ - \ cotA})^2$$ $$= \ ( \frac{1 \ - \ tanA}{ 1 \ - \ \frac{1}{tanA}})^2$$

$$= \ ( \frac{1 \ - \ tanA}{ \frac{tanA \ - \ 1}{tanA}})^2$$ $$= \ tan^2A$$