# If show that  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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