NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

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Written by Team Trustudies
Updated at 2021-05-07

NCERT solutions for class 10 Maths Chapter 8 Introduction To Trigonometry Exercise -8.1

Q1 ) In \(\triangle ABC \) , right angled at B, AB = 24 cm, BC = 7 cm. Determine :

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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

Given, in \( \triangle ABC \),

AB = 24 cm, BC = 7 cm and right angled at B

By using the Pythagoras theorem, we have
\(\Rightarrow AC^2 \ = \ AB^2 \ + \ BC^2 \)
\(\Rightarrow AC^2 \ = \ (24)^2 \ + \ (7)^2 \)
\( \Rightarrow AC^2 \ = \ 576 \ + \ 49 \ = \ 625 \)
\(\Rightarrow AC \ = \ \sqrt{625} \ = \ 25 \) cm

(i)
\( sinA \ = \ \frac{BC}{AC} \ = \ \frac{7}{25} \) [\(\because \) \( sin\theta \ = \ \frac{P}{H} \) ]
\( cosA \ = \ \frac{AB}{AC} \ = \ \frac{24}{25} \) [\(\because \)\( cos\theta \ = \ \frac{B}{H} \) ]

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

\( cosC \ = \ \frac{BC}{AC} \ = \ \frac{7}{25} \)

Q2 ) In Fig. find tan P – cot R.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

By using the Pythagoras theorem, we have

\( \Rightarrow PR^2 \ = \ PQ^2 \ + \ QR^2 \)
\(\Rightarrow 13^2 \ = \ 12^2 \ + \ QR^2 \)
\( \Rightarrow QR^2 \ = \ 13^2 \ - \ 12^2 \ \)
\( \Rightarrow QR^2 \ = \ 169 \ - \ 144 \ = \ 25 \)
\( \Rightarrow QR \ = \ \sqrt{25} \ = \ 5 \)

\(\therefore \) \( tanP \ = \ \frac{P}{B} \ = \ \frac{QR}{PQ} \ = \ \frac{5}{12} \)

\( cotR \ = \ \frac{B}{P} \ = \ \frac{QR}{PQ} \ = \ \frac{5}{12} \)

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

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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

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

Let BC = 3k and AC = 4k.

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

\( \therefore cosA \ = \ \frac{b}{h} \ = \ \frac{AB}{AC} \ = \ \frac{ \sqrt{7}k}{4k} \ = \ \frac{ \sqrt{7}}{4} \)

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

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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

Given,
\( 15 cot A \ = \ 8 \)
\( \Rightarrow cotA \ = \ \frac{8}{15} \ = \ \frac{B}{P} \ = \ \frac{AB}{BC} \)

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.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

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

Let AC = 13k and AB = 12k.

Then by Pythagoras theorem,

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

\( \therefore sin\theta \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{5k}{13k} \ = \ \frac{5}{13} \)

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

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

\( cot\theta \ = \ \frac{1}{tan\theta} \ = \ \frac{12}{5} \)

\( cosec\theta \ = \ \frac{1}{sin\theta} \ = \ \frac{13}{5} \)

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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

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

and, \( cosB \ = \ \frac{BC}{AB} \)

But given that, cos A = cos B

=> \( \frac{AC}{AB} \ = \ \frac{BC}{AB} \)
=> \( AC \ = \ BC \)

\(\therefore \) \( ∠ \ A \ = \ ∠ \ B \) [\(\because \) 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) \( cot^2\theta \)

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

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

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

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

\(\therefore \) \( sin\theta \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{8k}{ \sqrt{113}k} \ = \ \frac{8}{\sqrt{113}} \)

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.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

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

\(\therefore \) L.H.S = R.H.S.

\(\therefore \) \( \frac{1 - tan^2A}{1+tan^2A} \ = \ cos^2A \ - \ sin^2A \)

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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

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.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

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

Let QR = x cm
Thus, PR = (25 – x) cm

Then by Pythagoras theorem,

\(\Rightarrow RP^2 \ = \ RQ^2 \ + \ QP^2 \)

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

\( \Rightarrow 625\ - \ 50x \ + \ x^2 \ = \ x^2 \ + \ 25 \)

\( \Rightarrow - 50x \ = \ -600 \)

\( \Rightarrow x \ = \ \frac{600}{50} \ = \ 12 \)

\(\Rightarrow \) RQ = 12 cm

\(\Rightarrow \) 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 \( \theta \).

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

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

NCERT solutions for class 10 Maths Chapter 8 Introduction To Trigonometry 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) \( \frac{5cos^260° +4sec^230° - tan^245°}{sin^230° + cos^230°} \)

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

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.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

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°.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

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.

NCERT solutions for class 10 Maths Chapter 8 Introduction To Trigonometry Exercise -8.3

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

(ii) \( \frac{tan26°}{cot64°} \)

(iii) \( cos 48° – sin 42° \)

(iv) \( cosec 31° – sec 59°\)

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

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

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

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

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

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

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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

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

\( = \ \frac{1}{tan42°} × \frac{1}{tan67°} × tan42°tan67° \ = \ 1 \)

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

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

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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

Given,

\( tan 2A = cot (A – 18°) \)

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

=> \( 90° - 2A \ = \ A – 18° \) [ \(\because (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 \( A + B \ = \ 90° \).

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

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

=> \( cot B \ = \ tan(90° - B) \)

\( tan A \ = \ tan(90° - B) \)

=> \( (90° – A) \ = \ B \) [\(\because \) (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.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

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

=> \( cosec (90°– 4A) \ = \ sec 4A \)

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

=> \( 90° – 4A \ = \ A – 20° \) [\(\because \) (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
\( sin( \frac{B+C}{2}) \ = \ cos \frac{A}{2} \)

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

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°

\( \because A + B + C \ = \ 180° \)

=> \( \frac{A+B+C}{2} \ = \ 90° \)

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

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

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

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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

Given,

\( sin 67° + cos 75° \)

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

\( = \ cos 23° + sin 15° \)
[ \(\because \) \( sin( 90° - \theta) \ = \ cos \theta \) and \( cos( 90° - \theta) \ = \ sin \theta \) ]

NCERT solutions for class 10 Maths Chapter 8 Introduction To Trigonometry Exercise -8.4

Q1 ) Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

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.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

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

[ \(\because sin(90° - \theta) \ = \ cos\theta \)
\( \Rightarrow sin63° \ = \ sin(90° – 27°) \ = \ cos27° \) ]

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

\( = \ \frac{1}{1} \ = \ 1 \) [ \( \because cos^2A + sin^2A \ = \ 1 \) ]

(ii) \( sin25° cos65° + cos25°sin65° \)

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

= \( cos65° cos65° + sin 65° sin65° \) [\( \because sin(90°-\theta ) \ = \ cos\theta \) 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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

(i) (B), as

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

(ii) (C), as

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

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

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

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

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

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

(iii) (D), as

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

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

\( = \ \frac{1 \ - \ sin^2A}{cosA} \ = \ \frac{cos^2A}{cosA} \ = \ cosA \)
[ \( \because sin^2A \ + \ cos^2A \ = \ 1 \) ]

(iv) (D), as

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

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Answer :

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

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

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

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

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

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

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

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

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

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

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

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

\( = \ 1 \ + \ sec \theta cosec \theta \) = R.H.S.

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

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

\( \frac{sin^2A}{1 \ - \ cosA} \)
[ \( \because sin^2A \ = \ 1 \ - \ cos^2A \) ]

= R.H.S.

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

\( = \ \frac{cotA \ - \ 1 \ + \ cosecA}{cosA \ + \ 1 \ - \ cosecA} \)
[ \(\because 1 \ + \ cot^2A \ = \ cosec^2A \) ]

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

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 \( \sqrt{1+sinA} \) ]

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

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

\( = \ secA \ + \ tanA \)
[ \(\because tanA \ = \ \frac{sinA}{cosA} \) and \( secA \ = \ \frac{1}{cosA} \) ]

= R.H.S.

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

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

\( = \ tan\theta \)

= R.H.S.

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

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

\( = \ 1 \ + \ cot^2A \ + \ 1 \ + \ tan^2A \ + \ 5 \) [ \( \because sec^2A \ = \ 1 \ + \ tan^2A \) and \( cosec^2A \ = \ 1 \ + \ cot^2A \) ]

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

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

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

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

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. = \( ( \frac{1 \ + \ tan^2A}{1 \ + \ cot^2A} ) \)
\( = \ \frac{sec^2A}{cosec^2A} \)

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

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

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

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