Answer :


We know that,



Or



Let,


Base =BC = 12k


Hypotenuse =AC = 13k


Where, k ia any positive integer


So, by Pythagoras theorem, we can find the third side of a triangle


(AB)2 + (BC)2 = (AC)2


(AB)2 + (12k)2 = (13k)2


(AB)2 + 144k2 = 169k2


(AB)2 = 169 k2 –144 k2


(AB)2 = 25 k2


AB =25 k2


AB =±5k


But side AB can’t be negative. So, AB = 5k


Now, we have to find sin α and tan α


We know that,



Side opposite to angle α = AB =5k


And Hypotenuse = AC =13k


So,


We know that,



Side opposite to angle α = AB =5k


Side adjacent to angle α = BC =12k


So,


Now, LHS = sin α (1 – tan α)




= RHS


Hence Proved


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