# 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?

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:

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

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

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

You are almost done!

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

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.

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

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

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

\begin{align} \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{\sin A}{\cos A}\color{green}{\left(\cos A + 1\right)}}{\dfrac{1}{\sin A}\color{green}{(\cos A + 1)}}\\ &=\dfrac{\sin A}{\cos A} \div \dfrac{1}{\sin A}\\ &= \sin A \tan A \end{align}