Question

Simplify tan squared times (1 + cot squared x)

Answer 1

`\tan^2x\cdot(1+\cot^2x)\\\\\text{use}\ \tan x=(\sin x)/(\cos x)\ \text{and}\ \cot x=(\cos x)/(\sin x)\\\\=\left((\sin x)/(\cos x)\right)^2\cdot\left[1+\left((\cos x)/(\sin x)\right)^2\right]\\\\\text{use}\ \left((a)/(b)\right)^n=(a^n)/(b^n)\\\\=(\sin^2x)/(\cos^2x)\cdot\left(1+(\cos^2x)/(\sin^2x)\right)\\\\\text{use distributive property}\\\\=(\sin^2x)/(\cos^2x)\cdot1+(\sin^2x)/(\cos^2x)\cdot(\cos^2x)/(\sin^2x)`

`=(\sin^2x)/(\cos^2x)+1=\left((\sin x)/(\cos x)\right)^2+1=\tan^2x+1\\\\or\\\\=(\sin^2x)/(\cos^2x)+1=(\sin^2x)/(\cos^2x)+(\cos^2x)/(\cos^2x)=(\sin^2x+\cos^2x)/(\cos^2x)\\\\\text{use}\ \sin^2x+\cos^2x=1\\\\=(1)/(\cos^2x)=\left((1)/(\cos x)\right)^2\\\\\text{use}\ \sec x=(1)/(\cos x)\\\\=\sec^2x\\\\Answer:\ \tan^2x+1=(1)/(\cos^2x)=\sec^2x`

Answer 2

`\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)`