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,
Now, LHS = sin α (1 – tan α)
Hence Proved