# Class 11 RD Sharma Solutions – Chapter 5 Trigonometric Functions – Exercise 5.2

• Last Updated : 12 Mar, 2021

### (iv) sin x = 3/5, x in quadrant I

Solution:

(i) cot x = 12/5, x in quadrant III

As we knew that tan x and cot x are positive in third quadrant

and sin x, cos x, sec x, cosec x are negative.

By using the formulas,

tan x = 1/cot x

= 5/12

cosec x =

= -13/5

sin x = 1/cosec x

=- 5/13

cos x =

= -12/13

sec x = 1/cos x

= – 13/12

Hence, the values of the other five trigonometric functions are: 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

As we knew that sin x and cosec x are positive in second quadrant and

tan x, cot x, cos x, sec x are negative.

By using the formulas, we get

sin x =

= -2

Hence, the values of the other five trigonometric functions are: 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

As we knew that tan x and cot x are positive in third quadrant and sin x, cos x, sec x, cosec x are negative.

By using the formulas,

sin x =

Hence, the values of the other five trigonometric functions are: 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

As we knew that, all trigonometric ratios are positive in first quadrant.

So, by using the formulas,

tan x =

Hence, the values of the other five trigonometric functions are: 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.

Solution:

Given:

Sin x = and x lies in the second quadrant.

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

So, by using the formulas, we get

Cos x =

tan x = sin x/cos x

sec x = 1/cos x

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

= (-13 – 12)/5 = -25/5 = -5

Hence, the value of 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.

Solution:

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

Here, x is in second quadrant and y is in third quadrant. So, cos x and

tan x are negative in second quadrant and sec y is negative in third quadrant.

So, by using the formula, we get

cos x =

tan x = sin x/ cos x

cos x =

We know that sec y =

8tan x – √5 sec y = 8 × (-3)/(4) – √5 × (-√5/2) = -6 + (5/2) = (-12 + 5)/2 = -7/2

8tan x – √5 sec y = -7/2

Hence, the value of 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.

Solution:

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)

cos x and sec x are positive in fourth quadrant and

all other ratios are negative.

So, by using the formulas,

sec x =

cos x = 1/sec x

sin x =

sec x =

Hence, the value of sin x = -1/√2 and cos x = 1/√2

### Question 5. If cos x = -3/5 and π < x < 3π/2 find the values of other five trigonometric functions and hence evaluate

Solution:

Given, cos x = -3/5 and π <x < 3π/2

tan x and cot x are positive in the third quadrant and all other rations are negative.

Now, by using the formulas, we get

sin x = –

tan x = sin x/cos x

cot x = 1/tan x

sec x = 1/cos x

cosec x = 1/sin x

sin x =

tan x =

cot x =

sec x =

cosec x =

Now we evaluate:

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