`(csctheta-sintheta)(sectheta-costheta)=1/(tantheta+cottheta)`

To prove, let's simplify the left side and right side of the equation separately.

> For left side of the equation:

`(csctheta-sintheta)(sectheta-costheta)`

Note that `csctheta=1/(sin theta)` and `sectheta=1/(costheta)` .

`=(1/(sintheta)-sintheta)(1/(costheta)-costheta)= ((1-sin^2theta)/(sintheta))((1-cos^2theta)/(costheta))`

Then, apply the Pythagorean identity `sin^2 theta + cos^2 theta =1` .

`= ((cos^2 theta)/(sintheta))((sin^2theta)/(costheta)) = sinthetacostheta`

*So, the simplified form of left side of the equation is `sinthetacostheta` .*

> For the right side of the equation:

`1/(tan theta + cottheta)`

Note that `cot theta = 1/(tantheta)` .

`=1/(tantheta+1/(tan theta)) = 1/[(tan^2theta+1)/(tantheta)] = tan theta/(tan^2theta+1)`

Apply the Pythagorean identity `tan^2theta+1=sec^2theta` .

`= (tantheta)/(sec^2theta)`

Note that `tan theta =(sin theta)/(cos theta)` . Also, `sec theta = 1/(costheta)` .

`= [(sin theta)/(costheta)]/(1/(cos^2theta)) = (sintheta)/(cos theta)*(cos^2theta)/1`

`= sinthetacostheta`

The simplified form of right side of the equation is `sinthetacostheta` .

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**The simplified form of both sides of the equation is the same which proves the identity `(csctheta-sintheta)(sectheta-costheta)=1/(tantheta+cottheta)` .**

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