Verify the identify.\frac{\csc^2\theta}{1+\tan^2\theta}=\cot^2\theta

Ernstfalld 2021-09-01 Answered
Verify the identify.
\(\displaystyle{\frac{{{{\csc}^{{2}}\theta}}}{{{1}+{{\tan}^{{2}}\theta}}}}={{\cot}^{{2}}\theta}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

Mayme
Answered 2021-09-02 Author has 9455 answers
We know that
\(\displaystyle{\cot{{x}}}={\frac{{{\cos{{x}}}}}{{{\sin{{x}}}}}}\)
\(\displaystyle{\tan{{x}}}={\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}\)
\(\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1}\)
\(\displaystyle{L}{H}{S}={\frac{{{{\csc}^{{2}}\theta}}}{{{1}+{{\tan}^{{2}}\theta}}}}\)
\(\displaystyle={\frac{{{\frac{{{1}}}{{{{\sin}^{{2}}\theta}}}}}}{{{1}+{\frac{{{{\sin}^{{2}}\theta}}}{{{{\cos}^{{2}}\theta}}}}}}}\)
\(\displaystyle={\frac{{{\frac{{{1}}}{{{{\sin}^{{2}}\theta}}}}}}{{{\frac{{{{\cos}^{{2}}\theta}+{{\sin}^{{2}}\theta}}}{{{{\cos}^{{2}}\theta}}}}}}}\)
\(\displaystyle={\frac{{{\frac{{{1}}}{{{{\sin}^{{2}}\theta}}}}}}{{{\frac{{{1}}}{{{{\cos}^{{2}}\theta}}}}}}}\)
\(\displaystyle={\left({\frac{{{\cos{\theta}}}}{{{\sin{\theta}}}}}\right)}^{{2}}\)
\(\displaystyle={{\cot}^{{2}}\theta}\)
\(\displaystyle={R}{H}{S}\)
Have a similar question?
Ask An Expert
16
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question
...