[tex]\bf \textit{Pythagorean Identities} \\\\ sin^2(\theta)+cos^2(\theta)=1 \\\\ 1+cot^2(\theta)=csc^2(\theta) \\\\ 1+tan^2(\theta)=sec^2(\theta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ tan^2(x)[1+cot^2(x)]\implies tan^2(x)[csc^2(x)]\implies \cfrac{\underline{sin^2(x)}}{cos^2(x)}\cdot \cfrac{1}{\underline{sin^2(x)}} \\\\\\ \cfrac{1}{cos^2(x)}\implies sec^2(x)[/tex]
[tex]\tan^2x\cdot(1+\cot^2x)\\\\\text{use}\ \tan x=\dfrac{\sin x}{\cos x}\ \text{and}\ \cot x=\dfrac{\cos x}{\sin x}\\\\=\left(\dfrac{\sin x}{\cos x}\right)^2\cdot\left[1+\left(\dfrac{\cos x}{\sin x}\right)^2\right]\\\\\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\=\dfrac{\sin^2x}{\cos^2x}\cdot\left(1+\dfrac{\cos^2x}{\sin^2x}\right)\\\\\text{use distributive property}\\\\=\dfrac{\sin^2x}{\cos^2x}\cdot1+\dfrac{\sin^2x}{\cos^2x}\cdot\dfrac{\cos^2x}{\sin^2x}[/tex]
[tex]=\dfrac{\sin^2x}{\cos^2x}+1=\left(\dfrac{\sin x}{\cos x}\right)^2+1=\tan^2x+1\\\\or\\\\=\dfrac{\sin^2x}{\cos^2x}+1=\dfrac{\sin^2x}{\cos^2x}+\dfrac{\cos^2x}{\cos^2x}=\dfrac{\sin^2x+\cos^2x}{\cos^2x}\\\\\text{use}\ \sin^2x+\cos^2x=1\\\\=\dfrac{1}{\cos^2x}=\left(\dfrac{1}{\cos x}\right)^2\\\\\text{use}\ \sec x=\dfrac{1}{\cos x}\\\\=\sec^2x\\\\Answer:\ \tan^2x+1=\dfrac{1}{\cos^2x}=\sec^2x[/tex]
