# Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry – Exercise 8.4

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

Solution:

(i) sin A

We know that

cosec2A = 1 + cot2A

1/sin2A = 1 + cot2A

sin2A = 1/(1 + cot2A)

sin A = 1/(1+cot2A)1/2

(ii) sec A

sec2A = 1 + tan2A

Sec2A = 1 + 1/cot2A

sec2A = (cot2A + 1) / cot2A

sec A = (cot2A + 1)1/2 / cot A

(iii) tan A

tan A = 1 / cot A

tan A = cot -1 A

### Question 2. Write all the other trigonometric ratios of âˆ A in terms of sec A.

Solution:

(i) cos A

cos A = 1/sec A

(ii) sin A

We know that

sin2A = 1 – cos2A

Also , cos2A = 1 / sec2A

sin2A = 1 – 1 / sec2A

sin2A = (sec2A – 1) / sec2A

sin A = (sec2A – 1)1/2 / sec A

(iii) tan A

We know that

tan2A + 1 = sec2A

tan A = (sec2A – 1)Â½

(iv) cosec A

We know

cosec A = 1/ sinA

cosec A = sec A / (sec2A – 1)Â½

(v) cot A

We know

cot A = cos A / sin A

cot A = (1/sec A) / ((sec2A – 1)1/2 / sec A)

cot A = 1 / (sec2A – 1)1/2

### Question 3. Evaluate:

(i) (sin2 63Â° + sin2 27Â°)/(cos2 17Â° + cos2 73Â°)

(ii) sin 25Â° cos 65Â° + cos 25Â° sin 65Â°

(i) ([sin(90-27)]2 + sin2 27) / ([cos(90-73)]2 + cos2 73)

We know that

sin(90-x) = cos x
cos(90-x) = sin x

(cos2(27) + sin2 27) / (sin2(73) + cos2 73)

Using

sin2A + cos2A = 1

1/1 = 1

(ii) [sin 25 * cos(90-25)] + [cos 25 * sin(90-25)]

Using

sin(90-x) = cos x
cos(90-x) = sin x

= [sin 25 * sin 25] + [cos 25 * cos 25]

= sin2 25 + cos2 25

= 1

### Question 4. Choose the correct option. Justify your choice.

Solution:

(i) 9 sec2 A â€“ 9 tan2 A

(A) 1 (B) 9 (C) 8 (D) 0

Using sec2A – tan2A = 1

9 (sec2A – tan2A ) = 9(1)

Ans (B)

(ii) (1 + tan Î¸ + sec Î¸) (1 + cot Î¸ â€“ cosec Î¸)

(A) 0 (B) 1 (C) 2 (D) â€“1

Simplifying all ratios

= (1 + sinÎ¸/cosÎ¸ + 1/cosÎ¸) (1 + cosÎ¸/sinÎ¸ – 1/sinÎ¸)

= ((cosÎ¸ + sinÎ¸ + 1)/ cosÎ¸) ((sinÎ¸ + cosÎ¸ – 1 )/sinÎ¸)

= ((cosÎ¸ + sinÎ¸)2 – 1) / (sinÎ¸ cosÎ¸)

= (1 + 2*cosÎ¸*sinÎ¸ – 1) / (sinÎ¸ cosÎ¸)

= 2

Ans (C)

(iii) (sec A + tan A) * (1 â€“ sin A)

(A) sec A (B) sin A (C) cosec A (D) cos A

Simplifying sec A and tan A

= (1/cos A + sin A/cos A)*(1 – sin A)

= ((1 + sin A)/cos A)*(1 – sin A)

= (1 – sin2A)/cos A

= cos2A / cos A

= cos A

Ans (D)

(iv) (1 + tan2A) / (1 + cot2A)

(A) sec2A (B) â€“1 (C) cot2A (D) tan2A

Simplifying tan A and cot A

= (1 + (sin2A / cos2A)) / (1 + (cos2A / sin2A))

= ((cos2A + sin2A) / cos2A) / ((cos2A + sin2A) / sin2A)

= sin2A / cos2A

= tan2A

Ans (D)

### Question 5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Solution:

(i) (cosec Î¸ â€“ cot Î¸)2 = (1 – cosÎ¸) / (1 + cosÎ¸)

Solving LHS

Simplifying cosec Î¸ and cot Î¸

= (1-cos Î¸)2 / sin2Î¸

= (1-cos Î¸)2 / (1-cos2Î¸)

Using a2 – b2 = (a+b)*(a-b)

= (1-cos Î¸)2 / [(1-cos Î¸)*(1+cos Î¸)]

= (1-cos Î¸) / (1+cos Î¸) = RHS

Hence Proved

(ii) (cos A / (1+sin A) + ((1+sin A) / cos A) = 2 sec A

Solving LHS

Taking LCM

= (cos2A + (1+sin A)2) / ((1+sin A) cos A)

= (cos2A + 1 + sin2A + 2 sin A ) / ((1 + sin A)*cos A)

Using sin2A + cos2A = 1

= (2 + 2*sin A) / ((1+sin A)*cos A)

= (2*(1 + sin A)) / ((1 + sin A)*cos A)

= 2 / cos A

= 2 sec A = RHS

Hence Proved

(iii) (tan Î¸ / (1 – cot Î¸)) + (cot Î¸ / (1 – tan Î¸)) = 1 + sec Î¸*cosec Î¸

Solving LHS

Changing tan Î¸ and cot Î¸ in terms of sin Î¸ and cos Î¸ and simplifying

= ((sin2Î¸) / (cos Î¸ *(sin Î¸-cos Î¸))) + ((cos2Î¸ ) / (sin Î¸ *(sin Î¸-cos Î¸)))

= (1 / (sin Î¸-cos Î¸)) * [(sin3Î¸ – cos3Î¸) / (sin Î¸ * cos Î¸)]

= (1 / (sin Î¸ – cos Î¸)) * [ ((sin Î¸ – cos Î¸) * ( sin2Î¸ + cos2Î¸ + sin Î¸ * cos Î¸ ))/(sin Î¸ *cos Î¸)]

= (1+sin Î¸*cos Î¸) / (sin Î¸*cos Î¸)

= sec Î¸*cosec Î¸ + 1 = RHS

Hence Proved

(iv) (1 + sec A) / sec A = sin2A / (1 – cos A)

Solving LHS

= cos A + 1

Solving RHS

= (1 – cos2A) / (1 – cos A)

= (1 – cos A) * (1 + cos A) / (1 – cos A)

= 1 + cos A = RHS

Hence Proved

(v) (cos A – sin A + 1) / (cos A + sin A – 1) = cosec A + cot A using the identity cosec2A = 1 + cot2A

Solving LHS

Multiplying numerator and denominator by (cot A – 1 + cosec A)

= (cot2A + 1 + cosec2A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot2A – (1 + cosec2A – 2*cosec A))

= (2*cosec2A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot2A – 1 – cosec2A + 2*cosec A)

= (2* cosec A *(cosec A + cot A) – 2*(cosec A + cot A)) / (cot2A – 1 – cosec2A + 2*cosec A)

= ((cosec A + cot A) * (2*cosec A – 2 )) / (2*cosec A – 2)

= cosec A + cot A = RHS

Hence Proved

(vi) [(1 + sin A) / (1 – sin A)]Â½ = sec A + tan A

Solving LHS

Multiplying numerator and denominator by (1+sinA)

= [((1 + sin A)*(1 + sin A)) / ((1 – sin A)*(1 + sin A))]Â½

= (1 + sin A) / (1 – sin2A)Â½

= (1 + sin A) / (cos2A)1/2

= (1 + sin A) / (cos A)

= sec A + tan A = RHS

Hence Proved

(vii) (sin Î¸ – 2 sin3Î¸) / (2 cos3Î¸ – cos Î¸) = tan Î¸

Solving LHS

= (sin Î¸ * (1 – 2*sin2Î¸)) / (cos Î¸ * (2*cos2Î¸ – 1))

= (sin Î¸ * (1 – 2*sin2Î¸ )) / (cos Î¸ * (2*(1 – sin2Î¸) – 1))

= (sin Î¸ *(1 – 2*sin2Î¸)) / (cos Î¸ * (1 – 2*sin2Î¸))

= tan Î¸ = RHS

Hence Proved

(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A

Solving LHS

= sin2A + cosec2A + 2*sin A *cosec A + cos2A + sec2A + 2*cos A *sec A

We know that cosec A = 1 / sin A

= 1 + 1 + cot2A + 1 + tan2A + 2 + 2

= 7 + tan2A + cot2A = RHS

Hence Proved

(ix) (cosec A â€“ sin A)*(sec A â€“ cos A) = 1 / (tan A + cot A)

Solving LHS

= ((1/sin A) – sin A) * ((1/cos A) – cos A)

= ((1 – sin2A) / sin A) * ((1 – cos2A) / cos A)

= (cos2A * sin2A) / (sin A * cos A)

= sin A * cos A

Solving RHS

Simplifying tan A and cot A

= (sin A * cos A) / ( sin2A + cos2A)

= sin A * cos A = RHS

Hence Proved

(x) (1 + tan2A) / (1 + cot2A ) = [(1 – tan A) / (1 – cot A)]2 = tan2A

Solving LHS

Changing cot A = 1 / tan A

= (tan2A * (1 + tan2A)) / (1 + tan2A) = tan2A = RHS

= [(1 – tan A) / (1 – cot A)]2 = (1 + tan2A – 2*tan A) / (1 + cot2A – 2*cot A)

= (sec2A – 2*tan A) / (cosec2A – 2*cot A)

Solving this we get

= tan2A

Hence Proved

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