# Problem 8: Prove that cos^2x = (cos x*cosec x)/(cotx + tan x) (4)

*print*Print*list*Cite

### 3 Answers

The trigonometric identity `cos^2x = (cos x*cosec x)/(cotx + tan x)` has to be proved.

Start with the right hand side.

`(cos x*cosec x)/(cotx + tan x)`

Substitute `cot x = cos x/sin x` , `tan x = sin x/cos x` and `cosec x = 1/sin x`

= `(cos x*(1/sin x))/(cos x/sin x + sin x/cos x)`

= `(cos x/sin x)/((cos^2 x + sin^2x)/(sin x*cos x))`

= `(cos x)/((cos^2 x + sin^2x)/cos x)`

Use the property `sin^2 + cos^2 = 1` .

= `cos^2x`

**This proves that **`cos^2x = (cos x*cosec x)/(cotx + tan x)`

Note that cosec(x) can be express by csc(x)

The trigonometry identity:

Substitute` csc(x) =(1/sin(x))`

`=(cos(x)(1/sin(x)))/(cot(x)+tan(x))`

`=(cos(x)/sin(x))/(cot(x)+tan(x))`

Substitute `cot(x)= (cos(x)/sin(x))`

Then let `tan(x) =(1/cot(x))`

`= cot(x)/(cot(x)+(1/cot(x)))`

`=cot(x)/((cot^2(x)+1)/cot(x))`

Substitute `1+cot^2(x) =csc(x)`

`=cot(x)/((csc(x))/cot(x))`

Take the reciprocal of the divisor (bottom) and proceed to multiplication

`=cot(x)*(cot(x)/csc(x))`

`=(cot^2(x))/(csc^2(x))`

Substitute `cot^2(x) = (cos^2(x))/(sin^2(x)) ` and

`csc^2(x) = 1/(sin^2(x))`

`=((cos^2(x))/(sin^2(x)))/(1/(sin^2(x)))`

Cancel out `sin^2(x)`

=`(cos^2(x))/1 or cos^2(x)`

This proves `cos^2` =

Take note the last part is:

This proves that `cos^2(x) = (cos(x)csc(x))/(cot(x)+tan(x))`

``