Class 11 Maths Ncert Solutions Chapter 3 Ex 3.2 Trigonometric Functions PDF

Class 11 Maths Ncert Solutions Ex 3.2

Class 11 Maths Ncert Solutions Chapter 3 Ex 3.2

Q.1: Calculate the values of five trigonometric func. if cosy = 12 and y lies in 3rd quadrant.

Sol:

(i)  sec y :

Since, cos y = 12

Therefore, sec y = 1cosy = 1(12)

Hence, sec y = -2

 

(ii)  sin y :

Since, sin2y+cos2y=1

Therefore, sin2y=1cos2y

sin2y=1(12)2

sin2y=114

sin2y=34

siny=±32

Since, y lies in the third quadrant, the value of sin y will be negative.

Therefore, sin y = 32

 

(iii)  cosec y = 1siny = 1(32) = 23

Therefore, cosec y = 23

 

(iv)  tan y = sinycosy = tany=(32)(12) = 3

Therefore, tan y = 3

 

(v)  cot y = 1tany = 13

Therefore, cot y = 13

 

 

Q.2: Calculate the other five trigonometric function if we are given the values for sin y = 35, where y lies in second quadrant.

Sol:

sin y = 35

Therefore, cosec y = 1Siny = 135=53

Since, sin2y+cos2y=1

  cos2y=1sin2y

  cos2y=1(35)2

cos2y=1925

cos2y=1625

cos y = ±45

Since, y lies in the 2nd quadrant, the value of cos y will be negative,

Therefore, cos y = – 45

sec y = 1cosy=1(45)=54

tan y = sinycosy=(35)(45)=34

cot y = 1tany=43

 

 

 Q.3: Find the values of other five trigonometric functions if coty=34, where y lies in the third quadrant.

Sol:

cot y = 34

Since, tan y = 1coty=134=43

Since, 1+tan2y=sec2y

  1+(43)2=sec2y

  1+169=sec2y

  259=sec2y

  sec y = ±53

Since, y lies in the 3rd quadrant, the value of sec y will be negative.

Therefore, sec y = 53

cos y = 1secy=1(53)=35

Since, tan y = sinycosy

  43=siny(35)

sin y = (43)×(35)=45

  cosec y = 1siny=54

 

 

Q.4: Find the values of other five trigonometric if secy=135, where y lies in the fourth quadrant.

Sol:

sec y = 135

cos y = 1secy=1(135)=513

Since, sin2y+cos2y=1

   sin2y=1cos2y

   sin2y=1(513)2

   sin2y=125169=144169

sin y = ±1213

Since, y lies in the 4th quadrant, the value of sin y will be negative.

Therefore, sin y = 1213

cosec y = 1siny=1(1213)=1312

tan y = sinycosy=(1213)(513)=125

cot y = 1tany=1(125)=512

 

 

Q.5: Find the values of other five trigonometric function if tan y = 512 and y lies in second quadrant.

Sol:

tan  y = 512   [Given]

And, cot y = 1tany=1(512)=125

Since, 1+tan2y=sec2y

Therefore, 1+(512)2=sec2y

   sec2y=1+(25144)

  sec2y=169144

sec y = ±1312

Since, y lies in the 2nd quadrant, the value of sec y will be negative.

Therefore, sec  y = 1312

cos y = 1secy=1(1312)=(1213)

Since, tany=sinycosy

  512=siny(1213)

sin y = (512)×(1213)=513

cosec y = 1siny=1(513)=135

 

 

Q.6: Calculate the value of trigonometric function sin 765°.

Sol:

The values of sin y repeat after an interval of 360° or 2n.

Therefore, sin 765° = sin ( 2 × 360° + 45° ) = sin 45° = 12

 

 

Q.7: Calculate the value of trigonometric function cosec [-1410°]

 Sol:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, cosec [-1410°] = cosec [-1410° + 4 × 360°] = cosec [ -1410° + 1440°] = cosec 30° = 2

 

 

Q.8: Calculate the value of the trigonometric function tan19π3.

Sol:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, tan19π3 = tan613π = tan(6π+π3) = tanπ3 = tan 60° = 3

 

 

Q.9: Calculate the value of the trigonometric function sin(11π3).

 Answer:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, sin(11π3) = sin(11π3+2×2π) = sinπ3 = 32

 

 

Q.10: Calculate the value of the trigonometric function cot(15π4).

 Sol:

It is known that the values of tany repeat after an interval of 180 or n.

Therefore, cot(15π4) = cot(15π4+4π) = cotπ4 = 1

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