Prove 1 + tan²θ = sec²θ — From Pythagorean Identity

Question

Prove the identity: $1 + \tan^2\theta = \sec^2\theta$

This is a fundamental trigonometric identity — once you know it, a whole class of simplification problems becomes easy. It shows up in NCERT Class 10 and 11, and CBSE board papers ask students to either prove it or use it to simplify expressions.

Solution — Step by Step

Why This Works

The original identity $\sin^2\theta + \cos^2\theta = 1$ comes directly from Pythagoras’ theorem: in a right triangle with hypotenuse $r$, opposite side $y$, and adjacent side $x$, we have $x^2 + y^2 = r^2$. Dividing throughout by $r^2$ gives $\sin^2\theta + \cos^2\theta = 1$.

When we divide by $\cos^2\theta$, we’re essentially scaling the same geometric relationship differently — now measuring everything relative to the adjacent side instead of the hypotenuse. That’s why $\tan$ and $\sec$ (which are both defined relative to $\cos\theta$) appear naturally.

This “divide the master identity” trick generates a family of identities. Divide by $\sin^2\theta$ instead, and you get $1 + \cot^2\theta = \csc^2\theta$ — a CBSE favourite.

Alternative Method

Using the definition of sec and tan directly.

We know $\sec\theta = \dfrac{1}{\cos\theta}$ and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$.

So the right-hand side becomes:

This approach works backwards from RHS to LHS — sometimes CBSE questions explicitly ask you to “prove LHS = RHS” by working from one side only. This method satisfies that requirement cleanly.

Common Mistake