Question

$if \: \alpha - \beta = \frac{3\pi}{4} \: then \\ \\ (1 - \tan \: \alpha )(1 + \tan \: \alpha ) =$
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Answer 1

Given :-

• (A - B) = 3π/4.

To Find :-

• (1-tanA)(1+tanB) = ?

Solution :-

→ (A - B) = (3*180/4) = 135°

→ tan(A - B) = Tan135°

→ [ (tanA - tanB) / (1 + tanA*tanB) ] = Tan(90+45°)

→ [ (tanA - tanB) / (1 + tanA*tanB) ] = - tan45° = -1 .

→ (tanA - tanB) = - 1 - tanA*tanB

→ tanA - tanB + tanA*tanB = (-1)

→ tanA + tanA*tanB - tanB = (-1)

Adding (-1) both Sides now we get,

→ tanA + tanA*tanB - tanB - 1 = (-1) + (-1)

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

→ (1+tanB)(tanA - 1) = (-2)

Taking (-1) common,

→ (-1)(1+tanB)(1-tanA) = (-2)

→ (1-tanA)(1+tanB) = 2. (Ans).

Hence, Required Answer is 2.

Answer 2

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{ Question-}}}}}}$

$if \: \alpha - \beta = \frac{3\pi}{4} \: then \\ \\ (1 - \tan \: \alpha )(1 + \tan \: \alpha ) =$

Given = A-B =3π/4

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{ Solution-}}}}}}$

=> (A-B) = (3×180/4) =130°

=> tan(A-B) = tan135°

=> [ (tanA-tanB) / 1 + (tanA×tanB) ] = tan(90+45°)

=> [ (tanA -tanB) / (1+tanA×tanB) ] = -tan45° = -1

=> (tanA-tanB) = -1 - tanA × tanB

=> tanA - tanB + tanA × tanB = (-1)

Adding (-1) to both sides

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

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

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

Taking (-1) as common in the equations,

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

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

Thus,the answer obtained is 2