tan(x)sin(x) + sec(x)cos^2(x) = sec(x)

allhvasstH 2021-06-12 Answered
\(\displaystyle{\tan{{\left({x}\right)}}}{\sin{{\left({x}\right)}}}+{\sec{{\left({x}\right)}}}{{\cos}^{{2}}{\left({x}\right)}}={\sec{{\left({x}\right)}}}\)

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Expert Answer

Nicole Conner
Answered 2021-06-13 Author has 28611 answers

\(\displaystyle{\tan{{\left({x}\right)}}}=\frac{{\sin{{\left({x}\right)}}}}{{\cos{{\left({x}\right)}}}}.\)
\(\displaystyle{\tan{{\left({x}\right)}}}{\sin{{\left({x}\right)}}}=\frac{{{\sin}^{{2}}{\left({x}\right)}}}{{\cos{{\left({x}\right)}}}}.\)
\(\displaystyle{\sec{{\left({x}\right)}}}=\frac{{1}}{{\cos{{\left({x}\right)}}}};\)
\(\displaystyle{\sec{{\left({x}\right)}}}{{\cos}^{{2}}{\left({x}\right)}}={\cos{{\left({x}\right)}}}.\)
\(\tan(x)\sin(x) + \sec(x)\cos^2(x) = \frac{\sin^2(x)}{\cos(x)} + \cos(x)\)

\( = \frac{\sin^2(x) + \cos^2(x)}{\cos(x)} = \frac{1}{\cos(x)} = \sec(x) \{proved\}\)
as we know \(\displaystyle{{\sin}^{{2}}{\left({x}\right)}}+{{\cos}^{{2}}{\left({x}\right)}}={1}\).

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