`coth^2(x) - csc h^2(x) = 1` Verify the identity.

Monday, November 5, 2012

$\displaystyle{{\coth}}^{{2}}{\left({x}\right)}-{{\csc{{h}}}}^{{2}}{\left({x}\right)}={1}$ Verify the identity.

$\displaystyle{{\coth}}^{{2}}{\left({x}\right)}-{{\csc{{h}}}}^{{2}}{\left({x}\right)}={1}$

Take note that hyperbolic cotangent and hyperbolic cosecant are defined as

$\displaystyle{\coth{{\left({x}\right)}}}=\frac{{{{e}}^{{x}}+{{e}}^{{-{x}}}}}{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}$

$\displaystyle{{\csc{{h}}}}^{{2}}{\left({x}\right)}=\frac{{2}}{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}$

Plugging them, the left side of the equation becomes

$\displaystyle{{\left(\frac{{{{e}}^{{x}}+{{e}}^{{-{x}}}}}{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}\right)}}^{{2}}-{{\left(\frac{{2}}{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}\right)}}^{{2}}={1}$

$\displaystyle\frac{{{\left({{e}}^{{x}}+{{e}}^{{-{x}}}\right)}}^{{2}}}{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}-\frac{{{2}}^{{2}}}{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}={1}$

$\displaystyle\frac{{{\left({{e}}^{{x}}+{{e}}^{{-{x}}}\right)}}^{{2}}}{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}-\frac{{4}}{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}={1}$

$\displaystyle\frac{{{{\left({{e}}^{{x}}+{{e}}^{{-{x}}}\right)}}^{{2}}-{4}}}{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}={1}$

Then, simplify the numerator.

$\displaystyle\frac{{{\left({{e}}^{{x}}+{{e}}^{{-{x}}}\right)}{\left({{e}}^{{x}}+{{e}}^{{-{x}}}\right)}-{4}}}{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}={1}$

$\displaystyle\frac{{{{e}}^{{{2}{x}}}+{1}+{1}+{{e}}^{{-{2}{x}}}-{4}}}{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}={1}$

$\displaystyle\frac{{{{e}}^{{{2}{x}}}+{2}+{{e}}^{{-{2}{x}}}-{4}}}{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}={1}$

$\displaystyle\frac{{{{e}}^{{{2}{x}}}-{2}+{{e}}^{{-{2}{x}}}}}{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}={1}$

Factoring the numerator, it becomes

$\displaystyle\frac{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}}{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}={1}$

$\displaystyle\frac{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}={1}$

Cancelling common factor, the right side simplifies to

$\displaystyle{1}={1}$

This verifies that the given equation is an identity.

Therefore, $\displaystyle{{\coth}}^{{2}}{\left({x}\right)}-{{\csc{{h}}}}^{{2}}{\left({x}\right)}={1}$ is an identity.

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Monday, November 5, 2012

$\displaystyle{{\coth}}^{{2}}{\left({x}\right)}-{{\csc{{h}}}}^{{2}}{\left({x}\right)}={1}$ Verify the identity.

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Monday, November 5, 2012

Verify the identity.

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Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{{\coth}}^{{2}}{\left({x}\right)}-{{\csc{{h}}}}^{{2}}{\left({x}\right)}={1}$ Verify the identity.

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?