Trigonometric equations of rational form

Question

If $x=\frac{2 \sin A}{1+\cos A+\sin A}$ then what is the value of $\frac{\cos A}{1+\sin A}$ In the denominator I tried writin $1=\sin^2 A+\cos^2 A$. Then I am struck

Answer 1

Writing $A=2B,$

$$x=\frac{2\sin2B}{1+\cos2B+\sin2B}=\frac{2(2\sin B\cos B)}{1+2\cos^2B-1+2\sin B\cos B}=\frac{2\sin B}{\cos B+\sin B}$$

$$\implies\frac2x=\frac{\cos B+\sin B}{\sin B}=\cot B+1$$

Solve for $\cot B$

Now, $$\displaystyle\frac{\cos2B}{1+\sin 2B}=\frac{\cos^2B-\sin^2B}{(\cos B+\sin B)^2}=\frac{\cos B-\sin B}{\cos B+\sin B}=\frac{\cot B-1}{\cot B+1}$$

Answer 2

Since both expressions in the problem are rational functions of $\sin A$ and $\cos A$, the substitution $t=\tan\frac A2$ is useful. (See tangent half-angle substitution.) We get $$ x = \frac{2\sin A}{1+\cos A+\sin A} = \frac{2\left(\frac{2t}{1+t^2}\right)}{1 + \frac{1-t^2}{1+t^2} + \frac{2t}{1+t^2}} = \dots = 2 - \frac2{1+t} $$ On the other hand, $$ \frac{\cos A}{1+\sin A} = \frac{\left(\frac{1-t^2}{1+t^2}\right)}{1+\frac{2t}{1+t^2}} = \dots = \frac2{1+t} - 1 $$ Comparing these two yields $$ \frac{\cos A}{1+\sin A} = 1-x $$

Alternatively, a shorter but less methodical solution: \begin{align*} \frac{\cos A}{1+\sin A} &= \frac{(1+\sin A+\cos A)\cos A}{(1+\sin A+\cos A)(1+\sin A)} \\ &= \frac{(1+\sin A)\cos A+\cos^2 A}{(1+\sin A+\cos A)(1+\sin A)} \\ &= \frac{(1+\sin A)\cos A+(1-\sin^2 A)}{(1+\sin A+\cos A)(1+\sin A)} \\ &= \frac{\cos A+1-\sin A}{1+\sin A+\cos A} \\ &= 1 - \frac{2\sin A}{1+\sin A+\cos A} \\ \end{align*}