Prove the identity: (2 tan x − 2 sin x cos x) / (2 sin2 x) = tan x

Question

ScanSolve AI Tutor · Accepted Answer

To prove the identity, we will start with the left-hand side (LHS) and transform it into the right-hand side (RHS).

The identity to prove is:

\frac{2 \tan x - 2 \sin x \cos x}{2 \sin^2 x} = \tan x

Start with the Left Hand Side (LHS) and express $\tan x$ in terms of $\sin x$ and $\cos x$ .
$\text{LHS} = \frac{2 \left(\frac{\sin x}{\cos x}\right) - 2 \sin x \cos x}{2 \sin^2 x}$
Find a common denominator for the terms in the numerator.
$\text{LHS} = \frac{\frac{2 \sin x}{\cos x} - \frac{2 \sin x \cos x \cdot \cos x}{\cos x}}{2 \sin^2 x}$ $\text{LHS} = \frac{\frac{2 \sin x - 2 \sin x \cos^2 x}{\cos x}}{2 \sin^2 x}$
Factor out $2 \sin x$ from the numerator.
$\text{LHS} = \frac{\frac{2 \sin x (1 - \cos^2 x)}{\cos x}}{2 \sin^2 x}$
Apply the Pythagorean identity $1 - \cos^2 x = \sin^2 x$ .
$\text{LHS} = \frac{\frac{2 \sin x (\sin^2 x)}{\cos x}}{2 \sin^2 x}$ $\text{LHS} = \frac{\frac{2 \sin^3 x}{\cos x}}{2 \sin^2 x}$
Simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator.
$\text{LHS} = \frac{2 \sin^3 x}{\cos x} \cdot \frac{1}{2 \sin^2 x}$
Cancel out common terms (2 and $\sin^2 x$ ).
$\text{LHS} = \frac{\sin^3 x}{\sin^2 x \cos x}$ $\text{LHS} = \frac{\sin x}{\cos x}$
Recognize that $\frac{\sin x}{\cos x} = \tan x$ .
$\text{LHS} = \tan x$
Since LHS = RHS, the identity is proven.

$\text{The identity is proven.}$

Prove the identity: (2 tan x - 2 sin x cos x) / (2 sin2 x) = tan x