tan (45°-A) =1-tanA/1+tanA
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sorry I don't have any answer
Answered by
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Answer:
Step-by-step explanation:
Use tangent addition to prove that tan(A + 45) = (1 + tan(A))/(1 - tan(A))...
tan(α + ß) = (tanα + tanß)/(1 - tanαtanß)
So...
LHS
= tan(A + 45) = (tan(A) + tan(45))/(1 - tan(A)tan(45))
= (tan(A) + 1)/(1 - tan(A)(1))
= (1 + tan(A))/(1 - tan(A))
= RHS
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