tan( pie by 4 - theta) = 1-tan theta upon 1+tan theta proof .
Answers (2)
16th Sep, 2020
Hi student,
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!
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!
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Comments (0)
16th Sep, 2020
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.
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