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If two angles of a triangle are `tan^(-1)(2)a n dtan^(-1)(3),` then find the third angle.

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Given two angles are `tan^(-1) (2) and tan^(-1) (3)`. Now `(2) (3) gt 1`
`rArr tan^(-1) (2) and tan^(-1) (3) = pi + tan^(-1) ((2+3)/(1 - 2 xx 3))`
`= pi + tan^(-1) = pi - (pi)/(4) = (3pi)/(4)`
Hence, the third angle is `pi - (3pi)/(4) = (pi)/(4)`
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