3.7 IfA-B = = 3π/4, then S.T.(1-tana) (1 +tanB) =2.
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Step-by-step explanation:
Given, A-B = 3π/4
Taking tan on both sides,
tan(A-B) = tan(3π/4) = -1 ... (1)
Now using, tan(A-B) formula,
tan(A-B) = tan A - tan B
-----------------
1 + tan A * tan B ... (2)
Equating the right hand sides of (1) and (2)
tan A - tan B
----------------- = -1
1 + tan A * tan B
=> tan A - tan B = -1 - tanA*tanB
=> tanA - tanB + tanA*tanB = -1 ...(3)
Considering the LHS to prove:
=> (1-tanA) (1+tanB)
=> 1 - tanA + tanB - tanA*tanB
=> 1 - (tanA - tanB + tanA*tanB) ..(using equation 3)
=> 1 - (-1)
=> 2 = RHS
Hence, proved.
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