sinx = 9/25

First we will calculate the csecx.

We know that :

csecx= 1/sinx

==> csecx = 1/(9/25) = 25/9

Now let us calculate the cosx:

We know that:

sin^2 x + cos^2 x = 1

==? cos^2 x = 1 - sin^2 x

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

==> cosx = sqrt(1- (9/25)^2)

= sqrt(1- 81/626)

= sqrt(544)= 4sqrt(34)

==> cosx = 4sqrt(34)

==> secx = 1/cosx

==> secx = 1/4sqrt(34) = sqrt34/136

==> tanx = sinx/cosx = (9/25)/4sqrt34

==> tanx = 9/100sqrt34 = 9sqrt34/3400

Because we know, from enunciation, the value of the function sine, we could determine the value of the function cotan x.

cotan x = sqrt [1/(sin x)^2 - 1]

cotan x = [sqrt (25-9)(25+9)]/3

cotan x = sqrt(16)(34)/3

cotan x = 4sqrt 34/3

But tan x = 1/cotan x

tan x = 3/4sqrt34

tan x = 3sqrt34/4*34

**tan x = 3sqrt 34/ 136**

But tan x = sin x / cos x => cos x = sin x / tan x

cos x = (9/25)/(3sqrt 34/ 136)

cos x = 9*136/25*3*sqrt34

cos x = 3*136*sqrt 34/25*34

cos x = 3*68*sqrt 34/25*17

cos x = 3*4*sqrt 34/25

**cos x = 12*sqrt 34/25**

sec x = 1/ cos x

**sec x = 25*sqrt 34/12*34**

cosec x = 1/ sin x

**cosec x = 25/9**

sinx =9/25. To find cosx,cosex ,secx and tanx

We know that sinx is +ve in 1st and 2nd quadrant.

Also sin^2x+cos^2 = 1,by the fundamental trigonometric identity

So sinx = 9/25 implies

cosx = sqrt(1-sin^2x ) = + or- sqrt(1-9^2/25^2) = +or- 25 sqrt(25^2-9^2^2/25^2) = +or- sqrt(544)/25 = +or- (4sqrt34)/25 .

Cosx = 4sqrt34/25 in 1st quadrant and -4sqrt34/25 in 2ndd quadrant.

secx = 1/cosx = 25/(4sqrt34) in 1st quadrant and -25/(4sqrt34) in 2nd quadrant.

cosecx = 1/sinx = 1/(9/25) = 25/9 in 1st and 2nd quadrant

Tanx = sinx/cosx = (9/25)(25/4sqrt34) = 9/(3sqrt34) in 1st quadrant and -9/(4sqrt34) in 2nd quadrant.