Question

If A-B=3pi/4 then show that (1-tanA)(1+tanB)=2

Answer 1

Answer:

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

Step-by-step explanation:

A-B=3π/4

APPLY 'tan' ON BOTH SIDES

Tan(A-B)=Tan(3π/4)

tanA-tanB/1+tanAtanB=tan(135°)

tanA-tanB/1+tanAtanB= -1

tanA-tanB= -(1+tanAtanB)

tanB-tanA = 1-tanAtanB

tanB-tanA-tanAtanB = -1

ADD '-1' ON BOTH SIDES

-1+tanB-tanA-tanAtanB= -1+(-1)

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

                            Hence Proved

Answer 2

Given:

The difference between angles A and B is 3π/4.

To find:

The proof of the equation (1-tanA)(1+tanB) = 2.

Solution:

The given problem can be solved using the standard trigonometric formulae.

1. The difference between the angles A and B is,

=> A - B = 3π/4,

=> Apply Tan on both the sides,

=> Tan(A-B)=Tan(3π/4),

=> (TanA - TanB)/(1+TanATanB)=tan(135°),

=> (TanA-TanB)/(1+TanAtanB)= -1

=> (TanA-TanB)= -(1+TanATanB)

=> TanA - TanB = -1 -TanATanB,

=> TanA - Tan B + TanATanB = -1. (Consider as equation 1).

2. The LHS part of the equation to be proven is,

=>  (1-tanA)(1+tanB), ( Expand the expression ),

=> 1 + TanB -TanA -TanATanB,

=> 1 + - ( TanA + TanATanB - TanB ), (Value of ( TanA + TanATanB - TanB ) = -1 from equation 1 )

=> 1 + - ( -1 ),

=> 2 = RHS.

∴ Hence Proved

Therefore, the value of (1-tanA)(1+tanB) is 2.