I am unable to prove this Trigonometric Identity

Question

Can you please prove this identity:

$\displaystyle\frac{\cot A + \csc A - 1}{\cot A- \csc A + 1} = \frac{1+ \cos A}{\sin A}$

Answer 1

HINT:

Method $\#1:$

Set $\displaystyle1=\csc^2A-\cot^2A$ either in the numerator or in the denominator and try to take out the common factors.

Observe that $\displaystyle\frac{1+\cos A}{\sin A}=\csc A+\cot A$

Method $\#2:$

Multiply out the numerator & the denominator of the Left Side by $\sin A$

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

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

Now from the numerator,

$\displaystyle(\cos A+1-\sin A)\sin A=(1+\cos A)\sin A-\sin^2A$ $\displaystyle=(1+\cos A)\sin A-(1-\cos^2A)=(1+\cos A)(\sin A-1+\cos A)$

Method $\#3:$

Multiply out the numerator & the denominator of the Left Side by $1+\cos A$

Answer 2

Let's try writing the left side using $\sin A$ and $\cos A$ only, and then simplifying it to make it look like the right side. Using the fact that $$ \cos A = \frac{\cos A}{\sin A} \quad \mbox{and} \quad \csc A = \frac{1}{\sin A}$$ the left side turns into $$ \frac{\frac{\cos A}{\sin A} + \frac{1}{\sin A} - 1}{\frac{\cos A}{\sin A} - \frac{1}{\sin A} + 1}.$$ To simplify this further, let's try to make only one fraction on the top and bottom. We need a common denominator of $\sin A$ in both fractions. Writing $1$ as $\frac{\sin A}{\sin A}$ will give us $$ \frac{\frac{\cos A}{\sin A} + \frac{1}{\sin A} - \frac{\sin A}{\sin A}}{\frac{\cos A}{\sin A} - \frac{1}{\sin A} + \frac{\sin A}{\sin A}} = \frac{\frac{\cos A + 1 - \sin A}{ \sin A}}{\frac{\cos A - 1 + \sin A}{\sin A}} = \frac{\cos A + 1 - \sin A}{ \sin A} \cdot \frac{\sin A}{\cos A - 1 + \sin A}.$$ Cancelling out $\sin A$ and regroupin the terms will give us $$\frac{(\cos A - \sin A) + 1}{(\cos A + \sin A) - 1}.$$

From here, multiply the entire fraction by the conjugate of the bottom, which is $(\cos A + \sin A) + 1$. This will give you $$\frac{\cos^2 A - \sin^2 A + 2 \cos A + 1}{cos^2 A + 2\cos A \sin A + \sin^2 A - 1}.$$ Using the Pythagorean Identity $\cos^2 A + \sin^2 A = 1$ on the top and bottom (along with a little simplifying) will give you $\dfrac{1 + \cos A}{\sin A}$.