Show that $\dfrac{d}{dx}(\tan x)=\sec^2 x$ Show that $\sin 2x\cot x = 2\cos^2 x$ Hence, find the exact value of $\displaystyle\int_{\tfrac{\pi}{30}}^{\tfrac{\pi}{5}} \csc(10x)\tan(5x)\,dx$.

1. Express $\tan x$ as a quotient of sine and cosine, and set up the differentiation.

$\frac{d}{dx}(\tan x) = \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)$

2. Apply the quotient rule to differentiate the expression.

$= \frac{(\cos x)\frac{d}{dx}(\sin x) - (\sin x)\frac{d}{dx}(\cos x)}{(\cos x)^2}$
$= \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x}$

3. Simplify the numerator using the Pythagorean identity and conclude the proof using the reciprocal identity.

$= \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$
$= \frac{1}{\cos^2 x}$
$= \sec^2 x \text{ (shown)}$

1. Apply the double angle formula for sine and the ratio definition of cotangent

$\sin 2x \cot x = (2\sin x \cos x)\left(\frac{\cos x}{\sin x}\right)$

2. Simplify the expression to complete the proof

$= 2\cos x (\cos x)$\n$= 2\cos^2 x$

1. Rewrite the integrand using the identity from (b)

Write $\csc(10x)\tan(5x)=\dfrac{\tan(5x)}{\sin(10x)}=\dfrac{1}{\sin(10x)\cot(5x)}$.

Using (b) with $x\to 5x$: $\sin(10x)\cot(5x)=2\cos^2(5x)$.

Hence $\csc(10x)\tan(5x)=\dfrac{1}{2\cos^2(5x)}=\dfrac12\sec^2(5x).$

2. Integrate $\tfrac12\sec^2(5x)$

Using $\dfrac{d}{dx}\tan x=\sec^2 x$ from (a),
$\int \tfrac12\sec^2(5x)\,dx=\tfrac12\cdot\tfrac15\tan(5x)=\tfrac{1}{10}\tan(5x).$

3. Evaluate the definite integral

$\Big[\tfrac{1}{10}\tan(5x)\Big]_{\pi/30}^{\pi/5}=\tfrac{1}{10}\Big(\tan\pi-\tan\tfrac{\pi}{6}\Big)=\tfrac{1}{10}\Big(0-\tfrac{1}{\sqrt3}\Big)=-\tfrac{1}{10\sqrt3}=-\tfrac{\sqrt3}{30}.$