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If A and B are acute angles such that tan A =`1/3, tan B = 1/2 and tan (A + B) =` show that `A+B = 45^0`

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#### Solution

Given:

tan A = `1/3 and tan B = 1/2`

tan (A+B) = `(tan A + tan B)/(1- tan A tan B)`

On substituting these values in RHS of the expression, we get:

`(tan A + tan B )/(1- tan A tan B) = ((1/3 +1/2))/((1-1/3xx1/3)` =`((5/6))/(1-1/6) = ((5/6))/((5/6))=1`

⇒ tan (A + B) = 1= tan `45^0` [ ∵ tan 450 =1]

∴ A+B = 45^{0 }

Concept: Trigonometric Ratios and Its Reciprocal

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