Question

Transform cot B + tan B to csc B sec B

Trigonometric Functions
ANSWERED
asked 2021-05-13
Transform \(\displaystyle{\cot{{B}}}+{\tan{{B}}}\to{\csc{{B}}}{\sec{{B}}}\)

Answers (1)

2021-05-14
Use the quotient identities: \(\displaystyle{\cot{{B}}}+{\tan{{B}}}={\left(\frac{{\cos{{B}}}}{{\sin{{B}}}}\right)}+{\left(\frac{{\sin{{B}}}}{{\cos{{B}}}}\right)}\)
Add using the LCD which is \(\displaystyle{\sin{{B}}}{\cos{{B}}}\): \(\displaystyle{\cot{{B}}}+{\tan{{B}}}={\left(\frac{{\cos{{B}}}}{{\sin{{B}}}}\right)}{\left(\frac{{\cos{{B}}}}{{\cos{{B}}}}\right)}+{\left(\frac{{\sin{{B}}}}{{\cos{{B}}}}\right)}{\left({\sin{{B}}}+{\sin{{B}}}\right)}\)
\(\displaystyle{\cot{{B}}}+{\tan{{B}}}=\frac{{{{\cos}^{{2}}{B}}}}{{{\sin{{B}}}{\cos{{B}}}}}+\frac{{{\sin}^{{2}}{B}}}{{{\sin{{B}}}{\cos{{B}}}}}\)
\(\displaystyle{\cot{{B}}}+{\tan{{B}}}=\frac{{{{\cos}^{{2}}{B}}+{{\sin}^{{2}}{B}}}}{{\sin{{B}}}}{\cos{{B}}}\)
Use the Pythagorean identity: \(\displaystyle{{\cos}^{{2}}{B}}+{{\sin}^{{2}}{B}}={1}\)
\(\displaystyle{\cot{{B}}}+{\tan{{B}}}=\frac{{1}}{{\sin{{B}}}}{\cos{{B}}}\)
\(\displaystyle{\cot{{B}}}+{\tan{{B}}}={\left(\frac{{1}}{{\sin{{B}}}}\right)}{\left(\frac{{1}}{{\cos{{B}}}}\right)}\)
Use reciprocal identities:
\(\displaystyle{\cot{{B}}}+{\tan{{B}}}={\csc{{B}}}{\sec{{B}}}\)
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