# How does the trigonometric identity $1 + \cot^2\theta = \csc^2\theta$ derive from the identity $\sin^2\theta + \cos^2\theta = 1$?

I would like to understand how would the original identity of $$\sin^2 \theta + \cos^2 \theta = 1$$ derives into

$$1 + \cot^2 \theta = \csc^2 \theta$$

This is my working:

a) $$\frac{\sin^2 \theta}{ \sin^2 \theta } + \frac{\cos^2 \theta }{ \sin^2 \theta }= \frac 1 { \sin^2 \theta }$$

b) $$1 + \cot^2 \theta = \csc^2 \theta$$ How did the $\tan^2 \theta + 1 = \sec^2 \theta$ comes into the picture?

The same way; you start with $\sin^2\theta + \cos^2\theta = 1$ and divide both sides by $\cos^2\theta$.

• Ok, I got confused when the professor mentioned that we can derive the following result by dividing by sin^2 theta. He brought the tan^2 theta equation from the previous lecture. – ilovetolearn Jul 15 '17 at 3:55

It is taken from Pythagoras theorem, Let Know that let

c= Hypothenuse

a= Opposite

length of the triangle can be found by

$$a^2+b^2=c^2$$

Now divide everything by b

We have

$$\frac{a^2}{b^2}+1=\frac{c^2}{b^2}$$

Notice that

$$tan \theta=\frac{a}{b}$$

$$sec \theta=\frac{1}{\cos \theta}=\frac{c}{b}$$

There you go

$$1+\tan^2 \theta=\sec^2 \theta$$

There rest of the two identities are obtained by dividing c&a!

Which is Division by c $$\sin^2 \theta+\cos^2 \theta=1$$

Division by a

$$1+\cot^2\theta=\csc^2\theta$$

Take the original pythagorean identity, and divide by $\sin^2 x$ to get one of the identites, and divide by $\cos^2 x$ for the other.

Just as you divide $\sin^2(\theta)+\cos^2(\theta)=1$ by $\sin^2(\theta)$ to get $1 + \cot^2 (\theta) = \csc^2 (\theta)$,

you can divide $\sin^2(\theta)+\cos^2(\theta)=1$ by $\cos^2(\theta)$ to get $\tan^2 (\theta) + 1 = \sec^2(\theta)$