Problem 1 :
If A + B = 45°, show that
(1 + tan A)(1 + tan B) = 2
Solution :
A + B = 45°
tan (A + B) = tan 45°
(tan A + tan B)/(1 - tan A tan B) = 1
tan A + tan B = 1 - tan A tan B
Add tan A tan B on both sides.
tan A + tan B + tan A tan B = 1
tan A + tan B(1 + tan A) = 1
Add 1 on both sides
1 + tan A + tan B(1 + tan A) = 1 + 1
(1 + tan A) (1 + tan B) = 2
Hence proved.
Problem 2 :
Prove that (1 + tan1°)(1 + tan2°)(1 + tan3°).......(1 + tan44°) is a multiple of 4.
Solution :
= (1 + tan1°)(1 + tan2°)(1 + tan3°)................(1 + tan44°)
= (1 + tan1°)(1 + tan44°)(1 + tan2°)(1 + tan43°)(1 + tan3°)(1 + tan42°)................
If A + B = 45°, then
(1 + tanA)(1 + tanB) = 2
In the first two terms, we have 1 instead of A and 44 instead of B.
In a same manner, in third and 4th terms we have 2 instead of A and 43 instead of B. So, we get
= 2(2)(2)........................
It is a multiple of 4.
Problem 3 :
Prove that
tan (π/4 + θ) - tan (π/4 − θ) = 2 tan2θ
Solution :
tan(π/4 + θ) = (tanπ/4 + tanθ) / (1 - tanπ/4tanθ) :
tan(π/4 + θ) = (1 + tanθ) / (1 - tanθ)
tan(π/4 - θ) = (tanπ/4 - tanθ) / (1 + tanπ/4tanθ) :
tan(π/4 - θ) = (1 - tanθ) / (1 + tanθ)
Then,
tan(π/4 + θ) - tan(π/4 - θ) :
= [(1 + tanθ) / (1 - tanθ)] - [(1 - tanθ) / (1 + tanθ)]
= [(1 + tanθ)2 - (1 - tanθ)2] / (1 - tanθ)(1 + tanθ)
= [(1+tan2θ+2tanθ) - (1+tan2θ-2tanθ)]/(1-tanθ)(1+tanθ)
= 4tanθ/(1 - tan2θ)
= 2[2tanθ / (1 - tan2θ)]
= 2tan2θ
Hence proved.
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