[tex]if \: \alpha - \beta = \frac{3\pi}{4} \: then \\ \\ (1 - \tan \: \alpha )(1 + \tan \: \alpha ) = [/tex]
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[tex]if \: \alpha - \beta = \frac{3\pi}{4} \: then \\ \\ (1 - \tan \: \alpha )(1 + \tan \: \alpha ) = [/tex]
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Verified answer
Given :-
To Find :-
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.
[tex]\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{ Question-}}}}}} [/tex]
[tex]if \: \alpha - \beta = \frac{3\pi}{4} \: then \\ \\ (1 - \tan \: \alpha )(1 + \tan \: \alpha ) = [/tex]
Given = A-B =3π/4
[tex]\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{ Solution-}}}}}} [/tex]
=> (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