^{2}(x) + 1 = sec

^{2}(x)

We know that

sin

^{2}(x) + cos

^{2}(x) = 1

dividing both sides by cos

^{2}(x) we get

(sin

^{2}(x) + cos

^{2}(x))/cos

^{2}(x) = 1/cos

^{2}(x)

which equals

sin

^{2}(x)/cos

^{2}(x) + cos

^{2}(x)/cos

^{2}(x) = (1/cos(x))

^{2}

which can be re-written as

(sin(x)/(cos(x))

^{2}+ 1 = (1/cos(x))

^{2}

We know that tan

^{2}(x) = (sin(x)/(cos(x))

^{2}

and that (1/cos(x))

^{2}= sec

^{2}(x)

So we can write

cot

^{2}(x) + 1 = sec

^{2}(x)

## 2 comments:

there is a typo at the end you meant to say tan not cot

This is really an interesting topic. Congratulations to the writer. I'm sure a lot of readers having fun reading your post. Hoping to read more post from you in the future. Thank you and God bless!

www.imarksweb.org

Post a Comment