How do you verify the identity `tanx csc^2x-tanx=cotx` ?

Tuesday, May 20, 2014

How do you verify the identity $\displaystyle{\tan{{x}}}{{\csc}}^{{2}}{x}-{\tan{{x}}}={\cot{{x}}}$ ?

Hello!

Recall the definitions of tan, csc and cot:

$\displaystyle{\tan{{\left({x}\right)}}}=\frac{{\sin{{\left({x}\right)}}}}{{\cos{{\left({x}\right)}}}},$ $\displaystyle{\csc{{\left({x}\right)}}}=\frac{{1}}{{\sin{{\left({x}\right)}}}},{\cot{{\left({x}\right)}}}=\frac{{\cos{{\left({x}\right)}}}}{{\sin{{\left({x}\right)}}}}.$

Therefore the left part is equal to

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

$\displaystyle=\frac{{\sin{{\left({x}\right)}}}}{{\cos{{\left({x}\right)}}}}\cdot\frac{{{1}-{{\sin}}^{{2}}{\left({x}\right)}}}{{{{\sin}}^{{2}}{\left({x}\right)}}}=\frac{{\sin{{\left({x}\right)}}}}{{\cos{{\left({x}\right)}}}}\cdot\frac{{{{\cos}}^{{2}}{\left({x}\right)}}}{{{{\sin}}^{{2}}{\left({x}\right)}}}=\frac{{\cos{{\left({x}\right)}}}}{{\sin{{\left({x}\right)}}}},$

which is really equal to the right part, Q.E.D.

Hello!

Recall the definitions of tan, csc and cot:

Therefore the left part is equal to

which is really equal to the right part, Q.E.D.

at May 20, 2014

Notes

Tuesday, May 20, 2014

How do you verify the identity $\displaystyle{\tan{{x}}}{{\csc}}^{{2}}{x}-{\tan{{x}}}={\cot{{x}}}$ ?

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Tuesday, May 20, 2014