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

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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


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Answer :


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.



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Answer :


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.

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Answer :


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.

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Answer :


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.

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Answer :


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 \( ∠ \ A \ = \ ∠ \ B \).

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Answer :




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) \( cot^2\theta \)

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Answer :


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.

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Answer :


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) \( cos A cos C - sin A sin C \)

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Answer :


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.

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Answer :


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 \( \theta \).

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Answer :


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

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Answer :


(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

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Answer :


(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.

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Answer :


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

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Answer :


(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) \( cosec 31° – sec 59°\)

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Answer :


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

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Answer :


(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.

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Answer :

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 \( A + B \ = \ 90° \).

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Answer :

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.

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Answer :


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
\( sin( \frac{B+C}{2}) \ = \ cos \frac{A}{2} \)

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Answer :


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°.

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Answer :


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.

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Answer :


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.

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Answer :


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

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Answer :


(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

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Answer :


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

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Answer :


(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. = \( \frac{1 \ - \ tanA}{1 \ - \ cotA})^2 \) \( = \ ( \frac{1 \ - \ tanA}{ 1 \ - \ \frac{1}{tanA}})^2 \)

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