How does $1+ \tan^2x =\frac{1}{\cos^2x}$?

Question

I am unable to see why the following is true: $$1+ \tan^2x =\frac{1}{\cos^2x}$$ I have looked into the topic and I am familiar with the reciprocal ratios of cosec, sec, and cot. but cannot derive how this statement makes sense.

Any help on the topic would be very much appreciated.

Answer 1

If you already know that:

$\sin^2x+\cos^2x=1$

(And there are many videos showing this using a unit circle: https://m.youtube.com/watch?v=o-fAx_96lgw)

Then you should have no problem seeing:

$$\tan^2+1=\frac{1}{\cos^2x}$$

Because we just took the first equation and divided both sides by $\cos^2x$

Answer 2

Using $\tan(x)=\frac {\sin x}{\cos x}$ and the trigonometric identity you will be able to find the desired result $$1+\tan^2x=\frac {\cos^2x+\sin^2x}{\cos^2x}=\frac {1}{\cos^2x}$$

Answer 3

$$1+\tan^2(x)=\frac{1}{\cos^2(x)}\Longleftrightarrow$$ $$\cos^2(x)\left(1+\tan^2(x)\right)=1\Longleftrightarrow$$ $$\cos^2(x)\left(1+\left(\frac{\sin(x)}{\cos(x)}\right)^2\right)=1\Longleftrightarrow$$ $$\cos^2(x)\left(1+\frac{\sin^2(x)}{\cos^2(x)}\right)=1\Longleftrightarrow$$ $$\cos^2(x)+\frac{\cos^2(x)\sin^2(x)}{\cos^2(x)}=1\Longleftrightarrow$$ $$\cos^2(x)+\sin^2(x)=1$$

Answer 4

How, exactly, are you defining these functions? The very simplest is in terms of a right triangle in which "$cos(\theta)$" is defined as "length of near side" (leg closest to angle $\theta$) divided by the "length of the hypotenuse" and $tan(\theta)$ is defined as "length of opposite side divided by length of near side". If we let "a" represent the length of the "opposite side", b the length of the "near side", and c the length of the "hypotenuse", then the Pythagorean theorem says $a^2+ b^2= c^2$. Divideb^ both sides by $b^2$ to get $\frac{a^2}{b^2}+ 1= \frac{c^2}{b^2}$, $1+ \left(\frac{a}{b}\right)^2= \left(\frac{c}{b}\right)^2$ $1+ tan^2(\theta)= \frac{1}{cos^2(\theta)}$