Mathematics: Need with simplifying trig equations? i attempted them but i don t know how to complete them

The answer to this question isSquare root of 149...

Need with simplifying trig equations? i attempted

Ответ:

izzyisawesome5232

13.03.2022

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

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

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

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

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