Animated Solution for Mathematics - Trigonometry: Let $0<x<\pi /4$ then $(\sec 2x-\tan 2x)$ equals

We are tasked with simplifying the expression $E=\sec 2x-\tan 2x$ within the domain $0<x<\frac{\pi }{4}$.

Whenever you encounter secants, cosecants, or tangents, the most reliable strategy is to return to the fundamental definitions of sine and cosine. By rewriting the expression, we obtain:

Since the denominators are identical, we can combine these terms into a single, manageable fraction:

The presence of $1$ paired with a double-angle sine is a classic signal in JEE trigonometry. We utilize the Pythagorean identity $1={\cos }^{2}x+{\sin }^{2}x$ and the double-angle identity $\sin 2x=2\sin x\cos x$ to express everything in terms of the single angle $x$.

Substituting these into the numerator yields:

Recognizing the numerator as a perfect square, we simplify it to:

Next, we address the denominator using the identity $\cos 2x={\cos }^{2}x-{\sin }^{2}x$. This is a difference of squares, which factors into:

Given the domain $0<x<\frac{\pi }{4}$, we know that $\cos x>\sin x$. This ensures that the term $(\cos x-\sin x)$ is non-zero, allowing us to cancel it from the numerator and denominator:

To transform this into a tangent function, we divide both the numerator and the denominator by $\cos x$:

Recalling that $\tan (\frac{\pi }{4})=1$, we can rewrite the expression as:

Applying the compound angle formula for tangent, $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$, we arrive at the final simplified result: