Question

How do you write cos, tan, and sec in terms of csc?

Answer 1

`\bf \textit{Pythagorean Identities} \\ \quad \\ sin^2(\theta)+cos^2(\theta)=1 \\ \quad \\ 1+cot^2(\theta)=csc^2(\theta) \\ \quad \\ 1+tan^2(\theta)=sec^2(\theta)\\\\ -----------------------------\\\\ csc(\theta)=\cfrac{1}{sin(\theta)}\qquad \qquad sec(\theta)=\cfrac{1}{cos(\theta)}`

so hmm let us use those ones

then


`\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta) \\\\\\ cos(\theta)=√(1-sin^2(\theta))\implies cos(\theta)=\sqrt{1-(1)/(csc^2(\theta))} \\\\\\ cos(\theta)=\sqrt{\cfrac{csc^2(\theta)-1}{csc^2(\theta)}}\\\\ -----------------------------\\\\`


`\bf 1+cot^2(\theta)=csc^2(\theta)\implies 1+\cfrac{1}{tan^2(\theta)}=csc^2(\theta) \\\\\\ \cfrac{1}{tan^2(\theta)}=csc^2(\theta)-1\implies \cfrac{1}{csc^2(\theta)-1}=tan^2(\theta) \\\\\\ \sqrt{\cfrac{1}{csc^2(\theta)-1}}=tan(\theta)\\\\ -----------------------------\\\\ sec(\theta)=\cfrac{1}{cos(\theta)}\implies sec(\theta)=\cfrac{1}{\sqrt{(csc^2(\theta)-1)/(csc^2(\theta))}} \\\\\\ sec(\theta)=\sqrt{\cfrac{csc^2(\theta)}{csc^2(\theta)-1}}`

Answer 2

Prolly not the best person to answer this question lol