Question

Establish identity. \frac{\csc\theta-1}{\cot\theta}=\frac{\cot\theta}{\csc\theta+1}

Trigonometric equation and identitie
ANSWERED
asked 2021-09-10
Establish identity
\(\displaystyle{\frac{{{\csc{\theta}}-{1}}}{{{\cot{\theta}}}}}={\frac{{{\cot{\theta}}}}{{{\csc{\theta}}+{1}}}}\)

Expert Answers (1)

2021-09-11
Given identity \(\displaystyle{\frac{{{\csc{\theta}}-{1}}}{{{\cot{\theta}}}}}={\frac{{{\cot{\theta}}}}{{{\csc{\theta}}+{1}}}}\)
\(\displaystyle{\frac{{{\csc{\theta}}-{1}}}{{{\cot{\theta}}}}}={\frac{{{\cot{\theta}}}}{{{\csc{\theta}}+{1}}}}\)
First from the left-hand side,
\(\displaystyle{\frac{{{\csc{\theta}}-{1}}}{{{\cot{\theta}}}}}\)
Here use the trigonometry reciprocal identity, \(\displaystyle{\csc{\theta}}={\frac{{{1}}}{{{\sin{\theta}}}}}\) and \(\displaystyle{\cot{\theta}}={\frac{{{1}}}{{{\tan{\theta}}}}}\)
\(\displaystyle{\frac{{{\csc{\theta}}-{1}}}{{{\cot{\theta}}}}}={\frac{{{\frac{{{1}}}{{{\sin{\theta}}}}}-{1}}}{{{\frac{{{\cos{\theta}}}}{{{\sin{\theta}}}}}}}}\)
\(\displaystyle={\frac{{{\frac{{{1}-{\sin{\theta}}}}{{{\sin{\theta}}}}}}}{{{\frac{{{\cos{\theta}}}}{{{\sin{\theta}}}}}}}}\)
\(\displaystyle={\frac{{{1}-{\sin{\theta}}}}{{{\cos{\theta}}}}}\)
Multiply \(\displaystyle{\left({1}+{\sin{\theta}}\right)}\) in numerator and denominator,
\(\displaystyle{\frac{{{\csc{\theta}}-{1}}}{{{\cot{\theta}}}}}={\frac{{{\left({1}-{\sin{\theta}}\right)}{\left({1}+{\sin{\theta}}\right)}}}{{{\cos{\theta}}{\left({1}+{\sin{\theta}}\right)}}}}\)
\(\displaystyle={\frac{{{1}-{{\sin}^{{2}}\theta}}}{{{\cos{\theta}}{\left({1}+{\sin{\theta}}\right)}}}}\)
\(\displaystyle={\frac{{{{\cos}^{{2}}\theta}}}{{{\cos{\theta}}{\left({1}+{\sin{\theta}}\right)}}}}\)
\(\displaystyle={\frac{{{\cos{\theta}}}}{{{\left({1}+{\sin{\theta}}\right)}}}}\)
multiply \(\displaystyle{\frac{{{1}}}{{{\sin{\theta}}}}}\) in numerator and denominator,
\(\displaystyle{\frac{{{\csc{\theta}}-{1}}}{{{\cot{\theta}}}}}={\frac{{{\cos{\theta}}{\left({\frac{{{1}}}{{{\sin{\theta}}}}}\right)}}}{{{\left({1}+{\sin{\theta}}\right)}{\left({\frac{{{1}}}{{{\sin{\theta}}}}}\right)}}}}\)
\(\displaystyle={\frac{{{1}-{{\sin}^{{2}}\theta}}}{{{\cos{\theta}}{\left({1}+{\sin{\theta}}\right)}}}}\)
\(\displaystyle={\frac{{{{\cos}^{{2}}\theta}}}{{{\cos{\theta}}{\left({1}+{\sin{\theta}}\right)}}}}\)
\(\displaystyle={\frac{{{\cos{\theta}}}}{{{\left({1}+{\sin{\theta}}\right)}}}}\)
multiply \(\displaystyle{\frac{{{1}}}{{{\sin{\theta}}}}}\) in numerator and denominator,
\(\displaystyle{\frac{{{\csc{\theta}}-{1}}}{{{\cot{\theta}}}}}={\frac{{{\cos{\theta}}{\left({\frac{{{1}}}{{{\sin{\theta}}}}}\right)}}}{{{\left({1}+{\sin{\theta}}\right)}{\left({\frac{{{1}}}{{{\sin{\theta}}}}}\right)}}}}\)
\(\displaystyle={\frac{{{\cot{\theta}}}}{{{\csc{\theta}}+{1}}}}\)
Therefore it is established that the left-hand side equal to the right-hand side quantity.
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