Question

If A - B = $\frac{3\pi}{4}$, then show that (1 - tan A)(1 + tan B) = 2.

Answer 1

A - B = 3π/4

or, tan(A - B) = tan(3π/4)

or, (tanA - tanB)/(1 + tanA .tanB) = tan(π - π/4)

or, (tanA - tanB)/(1 + tanA. tanB) = -tan(π/4)

or, (tanA - tanB)/(1 + tanA. tanB) = -1

or, tanA - tanB = -1 - tanA. tanB

or, tanA - tanB + 1 + tanA. tanB = 0

or, tanA - tanB + 1 + tanA.tanB -2 = -2

or, tanA - 1 - tanB + tanA. tanB = -2

or, 1(tanA - 1) + tanB(tanA - 1) = -2

or, (tanA - 1)(1 + tanB) = -2

hence, (1 - tanA)(1 + tanB) = 2

Answer 2

HELLO DEAR,



GIVEN:- A - B = 3π/4

taking tangent both side,

=> tan(A - B) = tan(3π/4)

=> (tanA - tanB)/(1 + tanA .tanB) = tan(π - π/4)

=> (tanA - tanB)/(1 + tanA. tanB) = -tan(π/4)

=> (tanA - tanB)/(1 + tanA. tanB) = -1

=> tanA - tanB = -1 - tanA. tanB

=> tanA - tanB + 1 + tanA. tanB = 0

=> tanA - tanB + 1 + tanA.tanB -2 = -2

=> tanA - 1 - tanB + tanA. tanB = -2

=> 1(tanA - 1) + tanB(tanA - 1) = -2

=> (tanA - 1)(1 + tanB) = -2

Hence, (1 - tanA)(1 + tanB) = 2


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