# Verify trigonometry equation $\frac{\sin A+\tan A}{\cot A+\csc A}=\sin A \tan A$

Sorry for asking so many of these type of questions.

How would I verify the following trigonometry identity:

$$\frac{\sin A+\tan A}{\cot A+\csc A}=\sin A \tan A.$$

My work is

$$\frac{\sin A + \frac{\sin A}{\cos A}}{\frac{\cos A}{\sin A} + \frac{1}{\sin A}}.$$

Do I have to use a common denominator between the sin and tan to solve the identity?

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For what it's worth (at this late date)...

Since cotangent and cosecant are linked by the Pythagorean Identity, the "conjugate factor" method is helpful:

$$\frac{\sin A + \tan A}{\cot A + \csc A} \cdot \frac{\cot A - \csc A}{\cot A - \csc A} = \frac{(\sin A + \tan A)(\cot A - \csc A)}{\cot^2 A - \csc^2 A}$$

$$= \frac{\sin A \cot A - \sin A \csc A + \tan A \cot A - \tan A \csc A}{-1} = - (\cos A - 1 + 1 - \frac{1}{\cos A} )$$

$$= \frac{1 - \cos^2 A}{\cos A} = \frac{\sin^2 A}{\cos A} = \sin A \tan A .$$

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You are almost done!

$$\frac{\sin A+\tan A}{\cot A+\csc A}=\frac{\sin A+\tan A}{\frac1{\tan A}+\frac1{\sin A}}=\frac{\sin A+\tan A}{\frac{\sin A+\tan A}{\tan A\cdot\sin A}}=\tan A\cdot \sin A$$

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You are on the right track. You just need to simplify further. $$\dfrac{\sin(A) + \tan(A)}{\cot(A) + \csc(A)} = \dfrac{\sin(A) + \dfrac{\sin(A)}{\cos(A)}}{\dfrac{\cos(A)}{\sin(A)} + \dfrac1{\sin(A)}} = \dfrac{\sin(A) \left( \dfrac{1 + \cos(A)}{\cos(A)}\right)}{\dfrac{1+\cos(A)}{\sin(A)}}$$ Recall that $$\dfrac{a \times \dfrac{b}c}{\dfrac{d}{e}} = \dfrac{abe}{cd}$$ Hence, your expression simplifies to $$\dfrac{\sin(A) \left( \dfrac{1 + \cos(A)}{\cos(A)}\right)}{\dfrac{1+\cos(A)}{\sin(A)}} = \dfrac{\sin(A) (1+\cos(A)) \sin(A)}{\cos(A) (1+\cos(A))}$$ Cancel out $1+\cos(A)$ and use that $\dfrac{\sin(A)}{\cos(A)} = \tan(A)$, to get what you want.

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$$\dfrac {\sin A+\tan A}{\cot A+\csc A}\\=\dfrac{\sin A+\dfrac {\sin A}{\cos A}}{\dfrac{\cos A}{\sin A}+\dfrac{1}{sin A}}\\=\dfrac{\dfrac {\cos A\sin A+\sin A}{\cos A}}{\dfrac{\cos A+1}{\sin A}}\\=\dfrac{\color{red}{\sin A}(\cos A\sin A+\sin A)}{\color{red}{\cos A}(\cos A+1)}\\=\tan A\dfrac{\sin A\color{red}{(\cos A+1)}}{\color{red}{\cos A +1}}\\=\tan A \sin A$$

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+1 Clean and elegant – user22805 Jul 19 '12 at 9:29

Another approach is to think outside-of-the-box and just multiply top and bottom by $\sin A\tan A$. Notice that $\cot A\tan A = \sin A\csc A = 1$.

$$\frac{\sin A+\tan A}{\cot A+\csc A}=\frac{\sin A+\tan A}{\cot A+\csc A}\cdot\frac{\sin A\tan A}{\sin A\tan A}$$ $$=\frac{(\sin A+\tan A)(\sin A\tan A)}{\sin A\tan A \cot A + \csc A \sin A \tan A}$$ $$=\frac{(\sin A+\tan A)(\sin A\tan A)}{\sin A(\tan A \cot A) + (\csc A \sin A) \tan A}$$ $$=\frac{(\sin A+\tan A)(\sin A\tan A)}{\sin A + \tan A}$$ $$=\sin A\tan A$$

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