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