Question

tan(π/4+A)tan(π/4-A )=1​

Answer 1

1. tanpi/4+Atanpi/-A=(tanpi/4+tanA)/(1-tanpi/4tanA)×(tanpi/4-tanA)/(1+tanpi/4tanA)
2. 2..=(1+tanA)/(1-tanA)×(1-tanA)/(1+tanA)
3. =1

Answer 2

Question :-

$\bf \:Prove \: that \: tan(\dfrac{\pi}{4} + A)tan(\dfrac{\pi}{4} - A) = 1$

Solution:-

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Identity used :-

$\bf \:tan(x + y) = \dfrac{tanx + tany}{1 - tanx \: tany}$

$\bf \:tan(x - y) = \dfrac{tanx - tany}{1 + tanx \: tany}$

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$\bf \:Consider \: tan(\dfrac{\pi}{4} + A) \: tan(\dfrac{\pi}{4} - A)$

$\bf\implies \:\dfrac{tan\dfrac{\pi}{4} + tan A}{1 - tan\dfrac{\pi}{4} \times tan A} \times \dfrac{tan\dfrac{\pi}{4} - tan A}{1 + tan\dfrac{\pi}{4} \times tan A}$

$\bf\implies \:\dfrac{1 + tan A}{1 - tan A} \times \dfrac{1 - tan A}{1 + tan A}$

$\bf\implies \:1$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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