tan`theta+cottheta=secthetacsctheta`

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jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

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`

Sources:
aruv's profile pic

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

Since

`tan(theta)=(sin(theta))/(cos(theta))`

`therefore`

`cot(theta)=1/(tan(theta))=(cos(theta))/(sin(theta))`

`L.H.S.=tan(theta)+cot(theta)`

`=(sin(theta))/(cos(theta))+(cos(theta))/(sin(theta))`

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

`since`

`sin^2(theta)+cos^2(theta)=1`

`therefore`

`L.H.S.=1/(sin(theta)cos(theta))=1/(sin(theta))xx1/(cos(theta))`

`=cosec(theta)sec(theta)=sec(theta)cosec(theta)`

=R.H.S.

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