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...
See This Answer Now
Start your subscription to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Already a member? Log in here.
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`
Further Reading