# How do you prove the idenity `2sec theta=sec theta sin^2theta+cos theta `  ?can you do it step by step with an explanation of how youe doing it plz

lemjay | High School Teacher | (Level 2) Senior Educator

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`2 sec theta = sec theta sin^2theta + cos theta`

To prove, let's try to express right side in terms of secant function only. Note that `cos theta = 1/(sec theta)` . So,

`2 sec theta = sec theta sin^2 theta + 1/(sec theta)`

Then,express right side as one fraction. So multiply `sec theta sin^2 theta` by `(sec theta)/(sec theta)` .

`2 sec theta = sec theta sin^2 theta * (sec theta)/(sec theta) + 1/(sec theta)`

`2 sec theta= (sec^2 theta sin^2 theta)/(sec theta)+1/(sec theta)`

`2 sec theta= (sec^2 theta sin^2 theta + 1)/(sec theta)`

To simplify the numerator, apply the Pythagorean identity. So `sin^2 theta = 1 -cos^2 theta` .

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

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

Since `cos theta = 1/(sec theta)` , then `sec^2 theta*cos^2 theta = sec^2theta*1/(sec^2 theta) = 1` .

`2 sec theta = (sec^2 theta - 1 +1)/(sec theta)`

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

Then, cancel common factor between the numerator and denominator. Hence, expressing the right side in terms of secant function only yields:

`2 sec theta = sec theta `    (False)

And since the resulting expression at the right side is not the same with the expression at the left side, this proves that `2sec theta = sec theta sin^2theta + cos theta` is not an identity.