# Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry – Exercise 8.2

Last Updated : 13 Jan, 2021

### (i) sin 60Â° cos 30Â° + sin 30Â° cos 60Â°

Solution:

Formulas to be used : sin 30Â° = 1/2

cos 30Â° = âˆš3/2

sin 60Â° = 3/2

cos 60Â° = 1/2

=> (âˆš3/2) * (âˆš3/2) + (1/2) * (1/2)

=> 3/4 +1/4

=> 4 /4

=> 1

### (ii) 2 tan245Â° + cos230Â° â€“ sin260Â°

Solution:

Formulas to be used : sin 60Â° = âˆš3/2

cos 30Â° = âˆš3/2

tan 45Â° = 1

=> 2(1)(1) + (âˆš3/2)(âˆš3/2)-(âˆš3/2)(âˆš3/2)

=> 2 + 3/4 – 3/4

=> 2

### (iii) cos 45Â°/(sec 30Â°+cosec 30Â°)

Solution:

Formulas to be used : cos 45Â° = 1/âˆš2

sec 30Â° = 2/âˆš3

cosec 30Â° = 2

=> 1/âˆš2 / (2/âˆš3 + 2)

=> 1/âˆš2 / (2+2âˆš3)/âˆš3

=> âˆš3/âˆš2Ã—(2+2 âˆš3) = âˆš3/(2âˆš2+2âˆš6)

=> âˆš3(2âˆš6-2âˆš2)/(2âˆš6+2âˆš2)(2âˆš6-2âˆš2)

=> 2âˆš3(âˆš6-âˆš2) / (2âˆš6)Â²-(2âˆš2)Â²

=> 2âˆš3(âˆš6-âˆš2)/(24-8) = 2 âˆš3(âˆš6-âˆš2)/16

=> âˆš3(âˆš6-âˆš2)/8

=> (âˆš18-âˆš6)/8

=> (3âˆš2-âˆš6)/8

### (iv) (sin 30Â° + tan 45Âº – cosec 60Â°)/(sec 30Â° + cos 60Â° + cot 45Â°)

Solution:

Formulas to be used : sin 30Â° = 1/2

tan 45Â° = 1

cosec 60Â° = 2/âˆš3

sec 30Â° = 2/âˆš3

cos 60Â° = 1/2

cot 45Â° = 1

=> (1/2+1-2/âˆš3) / (2/âˆš3+1/2+1)

=> (3/2-2/âˆš3)/(3/2+2/âˆš3)

=> (3âˆš3-4/2 âˆš3)/(3âˆš3+4/2 âˆš3)

=> (3âˆš3-4)(3âˆš3-4)/(3âˆš3+4)(3âˆš3-4)

=> (27+16-24âˆš3) / (27-16)

=> (43-24âˆš3)/11

### (v) (5cos260Â° + 4sec2 30Â° – tan245Â°)/(sin230Â° + cosÂ²30Â°)

Solution:

Formulas to be used : cos 60Â° = 1/2

sec 30Â° = 2/âˆš3

tan 45Â° = 1

sin 30Â° = 1/2

cos 30Â° = âˆš3/2

=> 5(1/2)2+4(2/âˆš3)Â²-1Â²/(1/2)+(âˆš3/2)

=> (5/4+16/3-1) / (1/4+3/4)

=> (15+64-12) / 12/(4/4)

=> 67/12

### Question 2. Choose the correct option and justify your choice :

(i) 2tan 30Â°/1+tan230Â° =

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

(ii) 1-tan245Â°/1+tan245Â° =

(A) tan 90Â°            (B) 1                    (C) sin 45Â°            (D) 0

(iii)  sin 2A = 2 sin A is true when A =

(A) 0Â°                   (B) 30Â°                  (C) 45Â°                 (D) 60Â°

(iv) 2tan30Â°/1-tan230Â° =

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

Solution:

(i) In the given equation, substituting the value of tan 30Â°

As tan 30Â° = 1/âˆš3

2tan 30Â°/1+tan230Â° = 2(1/âˆš3)/1+(1/âˆš3)2

=> (2/âˆš3)/(1+1/3) = (2/âˆš3)/(4/3)

=> 6/4âˆš3 = âˆš3/2

=> sin 60Â°

The ans is sin 60Â°.

The correct option is (A).

(ii) In the given equation, substituting the of tan 45Â°

As tan 45Â° = 1

1-tan245Â°/1+tan245Â° = (1-12)/(1+12)

= 0/2 => 0

The ans is 0.

The correct option is (D).

(iii) sin 2A = 2 sin A is true when A = 0Â°

sin 2A = sin 0Â° = 0

2 sin A = 2 sin 0Â° = 2 Ã— 0 = 0

Another way :

sin 2A = 2sin A cos A

=> 2sin A cos A = 2 sin A

=> 2cos A = 2 => cos A = 1

Now, we have to check which degree value has to be applied, to get the solution as 1.

When 0 degree is applied to cos value we get 1, i.e., cos 0 = 1

Hence, A = 0Â°

The correct option is (A).

(iv) As tan 30Â° = 1/âˆš3

2tan30Â°/1-tan230Â° =  2(1/âˆš3)/1-(1/âˆš3)2

=> (2/âˆš3)/(1-1/3) = (2/âˆš3)/(2/3) = âˆš3 = tan 60Â°

The correct option is (C).

### Question 3. If tan (A + B) = âˆš3 and tan (A â€“ B) = 1/âˆš3, 0Â° < A + B â‰¤ 90Â°; A > B, find A and B.

Solution:

tan (A + B) = âˆš3

tan (A + B) = tan 60Â°

(A + B) = 60Â° â€¦ (i)

tan (A â€“ B) = 1/âˆš3

tan (A â€“ B) = tan 30Â°

(A â€“ B) = 30Â° â€¦  (ii)

Now add the equation (i) and (ii), we get

A + B + A â€“ B = 60Â° + 30Â°

A= 45Â°

Substituting the value of A in equation (i) to find the value of B

45Â° + B = 60Â°

B = 60Â° â€“ 45Â°

B = 15Â°

Hence, A = 45Â° and B = 15Â°

### Question 4. State whether the following are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B.

(ii) The value of sin Î¸ increases as Î¸ increases.

(iii) The value of cos Î¸ increases as Î¸ increases.

(iv) sin Î¸ = cos Î¸ for all values of Î¸.

(v) cot A is not defined for A = 0Â°.

Solution:

(i) Let us take A = 60Â° and B = 30Â°, then

Substitute the values of A and B in the sin (A + B) formula, we get

sin (A + B) = sin (60Â° + 30Â°) = sin 90Â° = 1 and,

sin A + sin B = sin 60Â° + sin 30Â°

= âˆš3/2 + 1/2 = (âˆš3 + 1 ) / 2, sin(A + B) â‰  sin A + sin B

Since both the values obtained are not equal.

Hence, the statement is false.

(ii) From the values given below, we can see that as angle(theta) increases value also increases.

sin 0Â° = 0, sin 30Â° = 1/2, sin 45Â° = 1/âˆš2, sin 60Â° = âˆš3/2 , sin 90Â° = 1

Thus, the value of sin Î¸ increases as Î¸ increases.

Hence, the statement is true.

(iii) From the values given below, we can see that as angle (theta) increases value decreases.

cos 0Â° = 1, cos 30Â° = âˆš3/2 , cos 45Â° = 1/âˆš2, cos 60Â° = 1/2, cos 90Â° = 0

Thus, the value of cos Î¸ decreases as Î¸ increases.

Hence, the statement given above is false.

(iv) sin Î¸ = cos Î¸, is only true for theta = 45Â°

Therefore, the above statement is false.

(v) As tan 0Â° = 0

cot 0Â° = 1 / tan 0Â°

= 1 / 0 => undefined

Hence, the given statement is true.

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