Question

1-tanA /1 +tanA =?
Solve the problem ​

Answer 1

Step-by-step explanation:

simplify the above equation accordingly and use accurate identities and u'll get the answer

Answer 2

Concept

The formula will use to solve the problem

1. $(a+b)^2=a^2+b^2+2ab$

2. $tanA=\frac{sinA}{cosA}$

3. $sin^2A+cos^2A=1$

Given

The expression is $\frac{1-tanA}{1 +tanA}$.

Find

We have to solve the problem.

Solution

Simplify the given expression

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

           $= \frac{\frac{cosA-sinA}{cosA}}{\frac{cosA+sinA}{cosA}}$

          $=\frac{cosA-sinA}{cosA+sinA}$

Multiply $cosA+sinA$ in numbeneter and denomenator

$\frac{cosA-sinA}{cosA+sinA}=\frac{(cosA-sinA)(cosA+sinA)}{(cosA+sinA)(cosA+sinA)}$

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

                $=\frac{cos^2A-sin^2A}{cos^2A+sin^2A+2cosAsinA}$

                $=\frac{cos2A}{1+sin2A}$

Hence the final value is $\frac{cos2A}{1+sin2A}$.