Verify the identity.\cot^2\theta-\cos^2\theta=\cot^2\theta\cos^\theta

Yasmin 2021-09-07 Answered
Verify the identity.
\(\displaystyle{{\cot}^{{2}}\theta}-{{\cos}^{{2}}\theta}={{\cot}^{{2}}\theta}{{\cos}^{\theta}}\)

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Expert Answer

Szeteib
Answered 2021-09-08 Author has 15850 answers
Given:
\(\displaystyle{{\cot}^{{2}}\theta}-{{\cos}^{{2}}\theta}={{\cot}^{{2}}\theta}{{\cos}^{\theta}}\)
Manipulating right side
\(\displaystyle{{\cot}^{{2}}\theta}{{\cos}^{{2}}\theta}\) Rewrite using trigonometry identities: \(\displaystyle{{\cos}^{{2}}\theta}={1}-{{\sin}^{{2}}\theta}\)
\(\displaystyle={\left({1}-{{\sin}^{{2}}\theta}\right)}{{\cot}^{{2}}\theta}\)
Expand \(\displaystyle{\left({1}-{{\sin}^{{2}}\theta}\right)}{{\cot}^{{2}}\theta}:{{\cot}^{{2}}\theta}-{{\cot}^{{2}}\theta}{{\sin}^{{2}}\theta}\)
\(\displaystyle={{\cot}^{{2}}\theta}-{{\cot}^{{2}}\theta}{{\sin}^{{2}}\theta}\)
Simplify using: \(\displaystyle{{\cot}^{{2}}\theta}={\frac{{{{\cos}^{{2}}\theta}}}{{{{\sin}^{{2}}\theta}}}}\)
\(\displaystyle={{\cot}^{{2}}\theta}-{\frac{{{{\cos}^{{2}}\theta}}}{{{\sin}^{\theta}}}}\times{{\sin}^{{2}}\theta}\)
\(\displaystyle={{\cot}^{{2}}\theta}-{{\cos}^{\theta}}\)
We showed that the two sides could take the same form.
Hence proved
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content_user
Answered 2021-12-10 Author has 11827 answers

Answer is given below (on video)

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