• +91 9971497814
• info@interviewmaterial.com

# RD Chapter 5- Trigonometric Functions Ex-5.2 Interview Questions Answers

### Related Subjects

Question 1 :
Find the values of the other five trigonometric functions in each of the following:
(i) cot x = 12/5, x in quadrant III
(ii) cos x = -1/2, x in quadrant II
(iii) tan x = 3/4, x in quadrant III
(iv) sin x = 3/5, x in quadrant I

(i) cot x = 12/5, x in quadrant III

In third quadrant, tan x and cot x are positive. sinx, cos x, sec x, cosec x are negative.

By using the formulas,

tan x = 1/cot x

= 1/(12/5)

= 5/12

cosec x = –(1 + cot2 x)

= –(1 + (12/5)2)

= –(25+144)/25

= –(169/25)

= -13/5

sin x = 1/cosec x

= 1/(-13/5)

= -5/13

cos x = – (1 – sin2 x)

= – (1 – (-5/13)2)

= – √(169-25)/169

= – √(144/169)

= -12/13

sec x = 1/cos x

= 1/(-12/13)

= -13/12

sin x = -5/13, cos x = -12/13,tan x = 5/12, cosec x = -13/5, sec x = -13/12

(ii) cos x = -1/2, x in quadrant II

In second quadrant, sin x and cosec x are positive.tan x, cot x, cos x, sec x are negative.

By using the formulas,

sin x = √(1 – cos2 x)

= √(1 – (-1/2)2)

= √(4-1)/4

= √(3/4)

= √3/2

tan x = sin x/cos x

= (√3/2)/(-1/2)

= -√3

cot x = 1/tan x

= 1/-√3

= -1/√3

cosec x = 1/sin x

= 1/(√3/2)

= 2/√3

sec x = 1/cos x

= 1/(-1/2)

= -2

sin x = √3/2, tan x =-√3, cosec x = 2/√3, cot x = -1/√3 sec x = -2

(iii) tan x = 3/4, x in quadrant III

In third quadrant, tan x and cot x are positive. sinx, cos x, sec x, cosec x are negative.

By using the formulas,

sin x = √(1 – cos2 x)

= – √(1-(-4/5)2)

= – √(25-16)/25

= – √(9/25)

= – 3/5

cos x = 1/sec x

= 1/(-5/4)

= -4/5

cot x = 1/tan x

= 1/(3/4)

= 4/3

cosec x = 1/sin x

= 1/(-3/5)

= -5/3

sec x = -√(1 + tan2 x)

= – √(1+(3/4)2)

= – √(16+9)/16

= – √ (25/16)

= -5/4

sin x = -3/5, cos x =-4/5, cosec x = -5/3, sec x = -5/4, cot x = 4/3

(iv) sin x = 3/5, x in quadrant I

In first quadrant, all trigonometric ratios arepositive.

So, by using the formulas,

tan x = sin x/cos x

= (3/5)/(4/5)

= 3/4

cosec x = 1/sin x

= 1/(3/5)

= 5/3

cos x = √(1-sin2 x)

= √(1 – (-3/5)2)

= √(25-9)/25

= √(16/25)

= 4/5

sec x = 1/cos x

= 1/(4/5)

= 5/4

cot x = 1/tan x

= 1/(3/4)

= 4/3

cos x = 4/5, tan x =3/4, cosec x = 5/3, sec x = 5/4, cot x = 4/3

Question 2 : If sin x = 12/13 and lies in the second quadrant, find the value of sec x + tan x.

Given:

Sin x = 12/13 and x lies in the second quadrant.

We know, in second quadrant, sin x and cosec x arepositive and all other ratios are negative.

By using the formulas,

Cos x = √(1-sin2 x)

= – √(1-(12/13)2)

= – √(1- (144/169))

= – √(169-144)/169

= -√(25/169)

= – 5/13

We know,

tan x = sin x/cos x

sec x = 1/cos x

Now,

tan x = (12/13)/(-5/13)

= -12/5

sec x = 1/(-5/13)

= -13/5

Sec x + tan x = -13/5 + (-12/5)

= (-13-12)/5

= -25/5

= -5

Sec x + tan x = -5

Question 3 : If sin x = 3/5, tan y = 1/2 and π/2 < x< π< y< 3π/2 find the value of 8 tan x -√5 sec y.

Given:

sin x = 3/5, tan y = 1/2 and π/2 < x< π

We know that, x is in second quadrant and y is inthird quadrant.

In second quadrant, cos x and tan x are negative.

In third quadrant, sec y is negative.

By using the formula,

cos x = – √(1-sin2 x)

tan x = sin x/cos x

Now,

cos x = – √(1-sin2 x)

= – √(1 – (3/5)2)

= – √(1 – 9/25)

= – √((25-9)/25)

= – √(16/25)

= – 4/5

tan x = sin x/cos x

= (3/5)/(-4/5)

= 3/5 × -5/4

= -3/4

We know that sec y = – √(1+tan2 y)

= – √(1 + (1/2)2)

= – √(1 + 1/4)

= – √((4+1)/4)

= – √(5/4)

= – √5/2

Now, 8 tan x – √5 sec y = 8(-3/4) – √5(-√5/2)

= -6 + 5/2

= (-12+5)/2

= -7/2

8 tan x – √5 sec y =-7/2

Question 4 : If sin x + cos x = 0 and x lies in the fourth quadrant, find sin x and cos x.

Given:

Sin x + cos x = 0 and x lies in fourth quadrant.

Sin x = -cos x

Sin x/cos x = -1

So, tan x = -1 (since, tan x = sin x/cos x)

We know that, in fourth quadrant, cos x and sec x arepositive and all other ratios are negative.

By using the formulas,

Sec x = √(1 + tan2 x)

Cos x = 1/sec x

Sin x = – √(1- cos2 x)

Now,

Sec x = √(1 + tan2 x)

= √(1 + (-1)2)

= √2

Cos x = 1/sec x

= 1/√2

Sin x = – √(1 – cos2 x)

= – √(1 – (1/√2)2)

= – √(1 – (1/2))

= – √((2-1)/2)

= – √(1/2)

= -1/√2

sin x = -1/√2 and cos x= 1/√2

Question 5 : If cos x = -3/5 and π Given:

cos x= -3/5 and π

We know that in the third quadrant, tan x and cot xare positive and all other rations are negative.

By using the formulas,

Sin x = – √(1-cos2 x)

Tan x = sin x/cos x

Cot x = 1/tan x

Sec x = 1/cos x

Cosec x = 1/sin x

Now,

Sin x = – √(1-cos2 x)

= – √(1-(-3/5)2)

= – √(1-9/25)

= – √((25-9)/25)

= – √(16/25)

= – 4/5

Tan x = sin x/cos x

= (-4/5)/(-3/5)

= -4/5 × -5/3

= 4/3

Cot x = 1/tan x

= 1/(4/3)

= 3/4

Sec x = 1/cos x

= 1/(-3/5)

= -5/3

Cosec x = 1/sin x

= 1/(-4/5)

= -5/4 = [(-5/4) + (3/4)] / [(-5/3) – (4/3)]

= [(-5+3)/4] / [(-5-4)/3]

= [-2/4] / [-9/3]

= [-1/2] / [-3]

= 1/6

Todays Deals  ### RD Chapter 5- Trigonometric Functions Ex-5.2 Contributors krishan

Name:
Email:

# Latest News # 9000 interview questions in different categories 