Prove that $ an 70^\circ - an 20^\circ = 2 an 50^\circ$

Question

Prove that $	an 70^\circ - 	an 20^\circ = 2	an 50^\circ$

Accepted Answer

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> [!TLDR]
> Combine the tangent difference into a single sine-over-cosine fraction, replace $\cos70^\circ$ by $\sin20^\circ$ (co-function), simplify with the double-angle formula, and the expression reduces exactly to $2	an50^\circ$.
>
> **Given:** Prove $	an 70^\circ - 	an 20^\circ = 2	an 50^\circ$.
>
> **Step 1.** Write both tangents with a common denominator:
> $$	an 70^\circ - 	an 20^\circ = \dfrac{\sin70^\circ}{\cos70^\circ} - \dfrac{\sin20^\circ}{\cos20^\circ} = \dfrac{\sin70^\circ\cos20^\circ - \cos70^\circ\sin20^\circ}{\cos70^\circ\cos20^\circ}$$
>
> **Step 2.** The numerator is $\sin(70^\circ - 20^\circ) = \sin 50^\circ$ (sine-subtraction formula):
> $$	an70^\circ - 	an20^\circ = \dfrac{\sin50^\circ}{\cos70^\circ\cos20^\circ}$$
>
> **Step 3.** Use $\cos70^\circ = \sin20^\circ$ (co-function identity), so the denominator becomes: …

Identity	Formula
Pythagorean	$\sin^2 \theta + \cos^2 \theta = 1$
Quotient	$\tan \theta = \frac{\sin \theta}{\cos \theta}$ , $\cot \theta = \frac{\cos \theta}{\sin \theta}$
Reciprocal	$\csc \theta = \frac{1}{\sin \theta}$ , $\sec \theta = \frac{1}{\cos \theta}$ , $\cot \theta = \frac{1}{\tan \theta}$
Even-Odd	$\sin(-\theta) = -\sin \theta$ , $\cos(-\theta) = \cos \theta$

Q.Prove that $\tan 70^\circ - \tan 20^\circ = 2\tan 50^\circ$

Concept understanding — Trigonometric Identity Proof

Trigonometric Identity Proof: From Intuition to Precision

The Core Idea

Proving an Identity: The Method

A Simple Example

The Toolbox of Known Identities

A Slightly Harder Proof

Q.Prove that tan⁡70∘−tan⁡20∘=2tan⁡50∘\tan 70^\circ - \tan 20^\circ = 2\tan 50^\circtan70∘−tan20∘=2tan50∘

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Q.Prove that $\tan 70^\circ - \tan 20^\circ = 2\tan 50^\circ$

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