tan( pie by 4 - theta) = 1-tan theta upon 1+tan theta proof .
We've to prove tan(pi/4 - x) = (1-tanx)/(1+tanx)
We know that tan(pi/4) i.e., tan(45deg) = 1
And tan(a-b) = ( tanA - tanB ) / ( 1+ tanA.tanB )
Let us prove by taking tan(pi/4 - x) and applying the formula, we get
tan(pi/4 - x) = tan(pi/4) - tanx) / ( 1 + tan(pi/4).tanx )
And when we substitute 1 at tan(pi/4) we get,
tan(pi/4 - x) = ( 1 - tanx )/ ( 1 + 1.tanx )
So that's why,
tan(pi/4 - x) = (1 - tanx)/(1 + tanx)
It's all about substitung the values and we get to prove the equation.
Hope this helps!
All the best!
Hello,
It is very simple to solve this problem. Let me show you.
R.H.S =1-tan theta/1+tan theta
=(Tan(π/4)-tan theta)/(1+tan(π/4)+tan theta)
= Tan(π/4-theta)
=L.H.S
Hence proved L.H.S=R.H.S
Hope this helps.