Prove that 2 tan inverse (1/3) + cot inverse 4 = tan inverse (16/13)

Solution :LHS = 2 tan-113 + cot-14=tan-12×131 - 19 + tan-114                   [as, cot-11x = tan-1x; when x > 0 and 2 tan-1x = tan-12x1-x2]=tan-12/38/9 + tan-114=tan-134 + tan-114=tan-13/4+1/41 - 34×14                                    [as, tan-1x + tan-1y = tan-1x+y1-xy]                 =tan-111-316=tan-11613=RHS

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