Simplify P(\sec \theta - \tan \theta)(1+\sin \theta)

opatovaL 2021-09-04 Answered
Simplify \(\displaystyle{\left({\sec{\theta}}-{\tan{\theta}}\right)}{\left({1}+{\sin{\theta}}\right)}\)

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Expert Answer

aprovard
Answered 2021-09-05 Author has 10808 answers
\(\displaystyle{\left({\sec{\theta}}-{\tan{\theta}}\right)}{\left({1}+{\sin{\theta}}\right)}=\)
\(\displaystyle={\sec{{\left(\theta\right)}}}{\sin{{\left(\theta\right)}}}-{\tan{{\left(\theta\right)}}}-{\tan{{\left(\theta\right)}}}{\sin{{\left(\theta\right)}}}=\)
\(\displaystyle={\frac{{{1}}}{{{\cos{\theta}}}}}+{\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}-{\tan{\theta}}-{\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}{\sin{\theta}}=\)
\(\displaystyle{\left[\because{\sec{\theta}}={\frac{{{1}}}{{{\cos{\theta}}}}}{\quad\text{and}\quad}{\tan{\theta}}={\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}\right]}\)
\(\displaystyle={\frac{{{1}}}{{{\cos{\theta}}}}}+{\tan{\theta}}-{\tan{\theta}}-{\frac{{{{\sin}^{{2}}\theta}}}{{{\cos{\theta}}}}}=\)
\(\displaystyle={\frac{{{1}}}{{{\cos{\theta}}}}}-{\frac{{{{\sin}^{{2}}\theta}}}{{{\cos{\theta}}}}}=\)
\(\displaystyle={\frac{{{1}-{{\sin}^{{2}}\theta}}}{{{\cos{\theta}}}}}=\)
\(\displaystyle={\frac{{{{\cos}^{{2}}\theta}}}{{{\cos{\theta}}}}}={\cos{\theta}}\)
\(\displaystyle{\left[\because{{\sin}^{{2}}\theta}+{{\cos}^{{2}}\theta}={1}\right]}\)
Answer: \(\displaystyle{\left({\sec{\theta}}-{\tan{\theta}}\right)}{\left({1}+{\sin{\theta}}\right)}={\cos{\theta}}\)
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