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In trigonometry we know that;

`tantheta = (sintheta)/(costheta)` `cottheta = 1/(tantheta) = (costheta)/(sintheta)`

`sectheta = 1/(costheta)`

`csctheta = 1/(sintheta)`

`tantheta+cottheta`

`= tantheta+1/(tantheta)`

`= (tantheta)^2/(tantheta)+1/(tantheta)`

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

We know that;

`1+tan^2theta = sec^2theta = 1/(cos^2theta)`

`tantheta+cottheta`

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

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

`= (costheta)/((cos^2thetaxxsintheta))`

`= 1/((costhetaxxsintheta))`

`= 1/(costheta)xx1/(sintheta)`

`= secthetaxxcsctheta`

So the answer...

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In trigonometry we know that;

`tantheta = (sintheta)/(costheta)`
`cottheta = 1/(tantheta) = (costheta)/(sintheta)`

`sectheta = 1/(costheta)`

`csctheta = 1/(sintheta)`

`tantheta+cottheta`

`= tantheta+1/(tantheta)`

`= (tantheta)^2/(tantheta)+1/(tantheta)`

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

We know that;

`1+tan^2theta = sec^2theta = 1/(cos^2theta)`

`tantheta+cottheta`

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

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

`= (costheta)/((cos^2thetaxxsintheta))`

`= 1/((costhetaxxsintheta))`

`= 1/(costheta)xx1/(sintheta)`

`= secthetaxxcsctheta`

So the answer is proved.

`tantheta+cottheta = secthetaxxcsctheta`

Approved by eNotes Editorial Team