Example Question - right-hand side

Proving a Trigonometric Identity

The given equation is a trigonometric identity that we need to prove. Let's start by working with the left-hand side (LHS) of the equation and try to transform it into the right-hand side (RHS). The LHS of the equation is: \[ \frac{2 \tan \theta}{1 + \tan^2 \theta} \] We know that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), and we can use this to rewrite the LHS: \[ \frac{2 \frac{\sin \theta}{\cos \theta}}{1 + \left(\frac{\sin \theta}{\cos \theta}\right)^2} \] Simplify it further by combining the terms: \[ \frac{2\sin \theta}{\cos \theta} \times \frac{\cos^2 \theta}{\cos^2 \theta + \sin^2 \theta} \] Since \(\cos^2 \theta + \sin^2 \theta = 1\) (which is another trigonometric identity), we can simplify: \[ \frac{2\sin \theta}{\cos \theta} \times \frac{\cos^2 \theta}{1} \] \[ 2\sin \theta \times \cos \theta \] Which is the RHS of the equation: \[ 2\sin \theta \cos \theta \] Therefore, we have proved that the LHS is equal to the RHS: \[ \frac{2 \tan \theta}{1 + \tan^2 \theta} = 2 \sin \theta \cos \theta \] And this confirms that the initial statement is a valid trigonometric identity.