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?
4 Answers
$\begingroup$It is taken from Pythagoras theorem, Let Know that let
c= Hypothenuse
a= Opposite
b=Adjacent
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
$$1+\cot^2\theta=\csc^2\theta$$
$\endgroup$ $\begingroup$The same way; you start with $\sin^2\theta + \cos^2\theta = 1$ and divide both sides by $\cos^2\theta$.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$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 $\tan^2 (\theta) + 1 = \sec^2(\theta)$
$\endgroup$
