How to simplify (cot^2 x)/(csc x +1)?
mamounette59cpg
Answered question
2023-01-20
Answer & Explanation
lilreeneyyh9
Beginner2023-01-21Added 7 answers
\(\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1}\)
\(\displaystyle\cancel{{\frac{{{\sin}^{{2}}{x}}}{{{\sin}^{{2}}{x}}}}}^{{{1}}}+{\stackrel{{{{\cot}^{{2}}{x}}}}{\overbrace{{\frac{{{\cos}^{{2}}{x}}}{{{\sin}^{{2}}{x}}}}}}}={\stackrel{{{{\csc}^{{2}}{x}}}}{\overbrace{{\frac{{1}}{{{\sin}^{{2}}{x}}}}}}}\)
\(\displaystyle{\mathbf{{{1}+{{\cot}^{{2}}{x}}={{\csc}^{{2}}{x}}}}}\)
Thus:
\(\displaystyle{\color{blue}{\frac{{{\cot}^{{2}}{x}}}{{{\csc{{x}}}+{1}}}}}\)
\(\displaystyle=\frac{{{{\csc}^{{2}}{x}}-{1}}}{{{\csc{{x}}}+{1}}}\)
\(\displaystyle=\frac{{{\left({\csc{{x}}}-{1}\right)}\cancel{{{\left({\csc{{x}}}+{1}\right)}}}}}{\cancel{{{\left({\csc{{x}}}+{1}\right)}}}}\)
\(\displaystyle={\color{blue}{{\csc{{x}}}-{1}}}\)
...if and only if \(\displaystyle{\csc{{x}}}+{1}\ne{0}\).
If \(\displaystyle{\csc{{x}}}+{1}={0}\), then \(\displaystyle\frac{{1}}{{\sin{{x}}}}=-{1}\), which is the case when \(\displaystyle{x}=\frac{{{3}\pi}}{{2}}\pm{2}{n}\pi\) for all \(\displaystyle{n}\in\mathbb{Z}\).
Hence, this answer is valid when \(\displaystyle{x}\ne\frac{{{3}\pi}}{{2}}\pm{2}{n}\pi\) for all \(\displaystyle{n}\in\mathbb{Z}\)."" What you're looking for is their identity is \(\displaystyle{{\csc}^{{2}}{x}}-{1}={{\cot}^{{2}}{x}}\).
You can derive this by starting from \(\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1}\):
\(\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1}\)
\(\displaystyle\cancel{{\frac{{{\sin}^{{2}}{x}}}{{{\sin}^{{2}}{x}}}}}^{{{1}}}+{\stackrel{{{{\cot}^{{2}}{x}}}}{\overbrace{{\frac{{{\cos}^{{2}}{x}}}{{{\sin}^{{2}}{x}}}}}}}={\stackrel{{{{\csc}^{{2}}{x}}}}{\overbrace{{\frac{{1}}{{{\sin}^{{2}}{x}}}}}}}\)
\(\displaystyle{\mathbf{{{1}+{{\cot}^{{2}}{x}}={{\csc}^{{2}}{x}}}}}\)
Thus:
\(\displaystyle{\color{blue}{\frac{{{\cot}^{{2}}{x}}}{{{\csc{{x}}}+{1}}}}}\)
\(\displaystyle=\frac{{{{\csc}^{{2}}{x}}-{1}}}{{{\csc{{x}}}+{1}}}\)
\(\displaystyle=\frac{{{\left({\csc{{x}}}-{1}\right)}\cancel{{{\left({\csc{{x}}}+{1}\right)}}}}}{\cancel{{{\left({\csc{{x}}}+{1}\right)}}}}\)
\(\displaystyle={\color{blue}{{\csc{{x}}}-{1}}}\)
...if and only if \(\displaystyle{\csc{{x}}}+{1}\ne{0}\).
If \(\displaystyle{\csc{{x}}}+{1}={0}\), then \(\displaystyle\frac{{1}}{{\sin{{x}}}}=-{1}\), which is the case when \(\displaystyle{x}=\frac{{{3}\pi}}{{2}}\pm{2}{n}\pi\) for all \(\displaystyle{n}\in\mathbb{Z}\).
Hence, this answer is valid when \(\displaystyle{x}\ne\frac{{{3}\pi}}{{2}}\pm{2}{n}\pi\) for all \(\displaystyle{n}\in\mathbb{Z}\)
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