# if sinx = 3/5 calculate cosx , tan, x and secx

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sinx = 3/5

We know that:

sin^2 x + cos^2 x = 1

==> cosx = sqrt(1-sin^2 x)

= sqrt(1- 9/25)

= sqrt( 16/25)

= 4/5

**==> cosx = 4/5**

Now we know that:

tanx = sinx/cosx = (3/5) / (4/5) = 3/4

**==> tanx = 3/4**

secx = 1/cosx = 1/(4/5) = 5/4

**==> secx = 5/4**

We know that (sin x )^2 + (cos x)^2 =1

Now sin x = 3/5

=> cos x = sqrt [ 1 - (sin x)^2]

=> cos x = sqrt [ 1 - (3/5)^2]

=> cos x = sqrt [ 1 - (9/25)]

=> cos x = sqrt ( 16 /25)

=> **cos x = 4/5**

**tan x **= sin x / cos x = (3/5) / (4/5) = **3/4**

**sec x **= 1/ tan x = **4/3**

Given sinx = 3/5.

cosx = sqrt(1-sin^2x) = sqrt(1-(3/5)^2) = 4/5.in tst quadrant or -4/5 in 2nd qudrant.

secx = 1/cosx = 5/4 in 1st and -5/4 in 2nd quadrant.

tanx = sinx/cosx = (3/5)/(4/5) = 3/4 in 1st and -3/4 in 2nd quadrant.