If tanA =tanP ,than prove that angleA =angle P

1.	If tanA =tanP ,than prove that angleA =angle P
Answer» Consider two right triangles XAY and WBZ such that tan A = tan BWe have,tan A =\xa0{tex}\\frac { X Y } { A Y }{/tex}\xa0and tan B =\xa0{tex}\\frac { WZ } { B Z }{/tex}Since, tan A = tan B{tex}\\Rightarrow \\frac { X Y } { A Y } = \\frac { W Z } { B Z }{/tex}{tex}\\Rightarrow \\frac { X Y } { W Z } = \\frac { A Y } { B Z } = k{/tex}(say)......(i){tex}\\Rightarrow X Y = k \\times W Z \\text { and } A Y = k \\times B Z{/tex}.......(ii)Using pythagoras\xa0theorem in triangles XAY and WBZ, we haveXA2\xa0= XY2\xa0+ AY2\xa0and WB2\xa0= WZ2\xa0+ BZ2{tex}\\Rightarrow{/tex}\xa0XA2\xa0= k2WZ2\xa0+ k2BZ2\xa0and WB2\xa0= WZ2\xa0+ BZ2{tex}\\Rightarrow{/tex}\xa0XA2\xa0= k2(WZ2\xa0+ BZ2) and WB2\xa0= WZ2\xa0+ BZ2{tex}\\Rightarrow\\frac { X A ^ { 2 } } { W B ^ { 2 } } = \\frac { k ^ { 2 } \\left( W Z ^ { 2 } + B Z ^ { 2 } \\right) } { \\left( W Z ^ { 2 } + B Z ^ { 2 } \\right) } = k ^ { 2 }{/tex}{tex}\\Rightarrow \\frac { X A } { W B } = k{/tex}...........(iii)From (i), (ii) and (iii), we get{tex}\\frac { X Y } { W Z } = \\frac { A Y } { B Z } = \\frac { X A } { W B }{/tex}{tex}\\Rightarrow \\Delta A Y X \\sim \\Delta B Z W{/tex}{tex}\\Rightarrow \\angle A = \\angle B{/tex}

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