Trig Identities

Question

Prove

sec²x = 1 + tan²x

^{I did R.H.S}

= 1 + sin²x / cos²x
= 1 + (1 - cos²x) / (1 - sin²x)
= 1 + cot²x
= ???

I think I'm doing it right, but I don't know how to solve now.

Answer 1

Hi Lawson,

Lets go back to the second line of what you have done so far:

= 1 + (1- cos2x)/(1 - sin2x)

 

We can continue from above by looking at the equation as the addition of a whole number to a fraction. The whole number is the 1 and the fraction is (1- cos2x)/(1 - sin2x). Therefore we first need to convert the whole number to a fraction with the same denominator:

 

=> 1 + (1- cos²x)/(1 - sin²x)

= (1 - sin²x)/(1 - sin²x) + (1- cos²x)/(1 - sin²x) = [(1 - sin²x) + (1 - cos²x)]/ (1 - sin²x)

 

Now if we replace all the 1s with cos² + sin²x, we will have:

 

[(cos²x + sin²x - sin²x) + (cos²x + sin²x - cos²x)]/ (cos²x + sin²x - sin²x)

= [cos²x + sin²x]/cos²x

= 1/cos²x

= sec²x

 

Hope this helps!

Answer 2

A better approach working with the RHS is to  write 

 

1 + sin²x/cos²x  =        (cos²x + sin²x)/cos²x  =  1/ cos²x  =  sec²x

 

The first step is essentially adding fractions, the second used the Pythagorean identity.