Question

Need help with simplifying trig equations? I attempted them but I don't know how to complete them

Answer 1

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


`\bf \cfrac{\boxed{tan^2(\theta)+1}+2tan(\theta)-sec^2(\theta)}{tan(\theta)} \\\\\\ \cfrac{\underline{\boxed{sec^2(\theta)}}+2tan(\theta)\underline{-sec^2(\theta)}}{tan(\theta)}\implies \cfrac{2tan(\theta)}{tan(\theta)}\implies 2\\\\ -------------------------------\\\\ 2)\\\\ \cfrac{5cos^2(\theta)+6(\theta)+1}{cos^2(\theta)-1}\implies \cfrac{[5cos(\theta)+1]\underline{[cos(\theta)+1]}}{[cos(\theta)-1]\underline{[cos(\theta)+1]}} \\\\\\ \cfrac{5cos(\theta)+1}{cos(\theta)-1}`


`\bf -------------------------------\\\\ 3)\\\\\ [1+cot(\theta )][1-cot(\theta )]-csc^2(\theta )\implies [1^2-cot^2(\theta )]-csc^2(\theta ) \\\\\\ \boxed{1-csc^2(\theta )}-cot^2(\theta )\implies \boxed{-cot^2(\theta )}-cot^2(\theta )\implies -2cot^2(\theta )`