Question

√1-cos A/ 1+ cos A = sinA/1+cosA​

Answer 1

Step-by-step explanation:

Show that;       $\sqrt{\frac{1-CosA}{1+CosA} } =\frac{SinA}{1+CosA}$

First to prove the following we need to multiply and divide $\sqrt{1+CosA}$ with $\sqrt{\frac{1-CosA}{1+CosA} }$

That implies;   $\sqrt{\frac{(1-CosA)(1+CosA)}{(1+CosA)(1+CosA)} }$

                   $=\sqrt{\frac{1-Cos^2A}{(1+CosA)^2} }$      We know that: $1-Cos^2A=Sin^2A$

                  $=\sqrt{\frac{Sin^2A}{(1+Cos^2A)} }$

                  $=\frac{SinA}{1+CosA}$

Hence Proved;

Answer 2

LHS

$= \sqrt{ \frac{1 - cosA}{1 + cosA} } \\ \\ = \sqrt{ \frac{1 - cosA}{1 + cosA} \times \frac{1 + cosA}{1 + cosA} } \\ \\ = \sqrt{ \frac{1 - {cos}^{2}A }{( {1 + cosA})^{2} } } \\ \\ = \sqrt{ \frac{ {sin}^{2} A}{( {1 + cosA})^{2} } } \\ \\ = \frac{sinA}{1 + cosA}$

RHS