Derive $(1+ \tan^2x= \sec^2x )$ and $(1+\cot^2x = \text{cosec}^2x)$ geometrically.

Question

This a circle through which you have to derive the above given identity.

Answer 1

As @Ak19 hinted, Pythagoras gives $\sin^2x+\cos^2x=1$. Divide by $\cos^2x$ or $\sin^2x$.

Answer 2

Consider this, looking at the image you have provided $\cos x = a$ because the cosine of an angle is the side of the triangle adjacent to it divided by the hypotenuse (longest side).

Given the radius of the circle is $= 1, \cos x = a/1 = a$ using the similar identity of $\sin$ we come to the conclusion that $\sin x = b$.

So as mentioned by the comments and previous answer using the Pythagoras theorem the sum of squaring both a and b should equal to 1 as that is the longest side of the triangle. (the a and b we worked out in the previous step using the identities of sine and cosine)

Now that we have arrived at the conclusion that $\sin^2x+\cos^2x=1$, by dividing the entire equation by $(\cos x)^2$ or $(\sin x)^2$ gives the results $1+\tan^2x=\sec^2x$ and $1+\cot^2x=\operatorname{cosec}^2x$

Answer 3

Geometrically, use similar right triangles:

Answer 4

Without words (use Pythagoras).