Prove that `2 + sec(x) cosec(x) = (sin x + cos x)^2 / (sin x cos x).`

Saturday, June 4, 2011

Prove that $\displaystyle{2}+{\sec{{\left({x}\right)}}}{\cos{{e}}}{c}{\left({x}\right)}=\frac{{{\left({\sin{{x}}}+{\cos{{x}}}\right)}}^{{2}}}{{{\sin{{x}}}{\cos{{x}}}}}.$

$\displaystyle{2}+{\sec{{x}}}{\csc{{x}}}=\frac{{{\left({\sin{{x}}}+{\cos{{x}}}\right)}}^{{2}}}{{{\sin{{x}}}{\cos{{x}}}}}$

To prove, consider the left side of the equation.

$\displaystyle{2}+{\sec{{x}}}{\csc{{x}}}$

Express the secant and cosecant in terms of cosine and sine, respectively.

$\displaystyle={2}+\frac{{1}}{{\cos{{x}}}}\cdot\frac{{1}}{{\sin{{x}}}}$

$\displaystyle={2}+\frac{{1}}{{{\sin{{x}}}{\cos{{x}}}}}$

To add, express them as two fractions with same denominators.

$\displaystyle={2}\cdot\frac{{{\sin{{x}}}{\cos{{x}}}}}{{{\sin{{x}}}{\cos{{x}}}}}+\frac{{1}}{{{\sin{{x}}}{\cos{{x}}}}}$

$\displaystyle=\frac{{{2}{\sin{{x}}}{\cos{{x}}}}}{{{\sin{{x}}}{\cos{{x}}}}}+\frac{{1}}{{{\sin{{x}}}{\cos{{x}}}}}$

$\displaystyle=\frac{{{2}{\sin{{x}}}{\cos{{x}}}+{1}}}{{{\sin{{x}}}{\cos{{x}}}}}$

Apply the Pythagorean identity $\displaystyle{{\sin}}^{{2}}{x}+{{\cos}}^{{2}}{x}={1}$ .

$\displaystyle=\frac{{{2}{\sin{{x}}}{\cos{{x}}}+{{\sin}}^{{2}}+{{\cos}}^{{2}}{x}}}{{{\sin{{x}}}{\cos{{x}}}}}$

$\displaystyle=\frac{{{{\sin}}^{{2}}{x}+{2}{\sin{{x}}}{\cos{{x}}}+{{\cos}}^{{2}}{x}}}{{{\sin{{x}}}{\cos{{x}}}}}$

And, factor the numerator.

$\displaystyle=\frac{{{\left({\sin{{x}}}+{\cos{{x}}}\right)}{\left({\sin{{x}}}+{\cos{{x}}}\right)}}}{{{\sin{{x}}}{\cos{{x}}}}}$

$\displaystyle=\frac{{{\left({\sin{{x}}}+{\cos{{x}}}\right)}}^{{2}}}{{{\sin{{x}}}{\cos{{x}}}}}$

Notice that this is the same expression that the right side of the equation have. Thus, this proves that the $\displaystyle{2}+{\sec{{x}}}{\csc{{x}}}=\frac{{{\left({\sin{{x}}}+{\cos{{x}}}\right)}}^{{2}}}{{{\sin{{x}}}{\cos{{x}}}}}$ is an identity.

Ninenotes

Saturday, June 4, 2011

Prove that $\displaystyle{2}+{\sec{{\left({x}\right)}}}{\cos{{e}}}{c}{\left({x}\right)}=\frac{{{\left({\sin{{x}}}+{\cos{{x}}}\right)}}^{{2}}}{{{\sin{{x}}}{\cos{{x}}}}}.$

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