If , prove that q sin θ = p.

Given : q cos θ = √(q² – p²)

We know that,

Or

Let,

Base =BC = √(q² – p²)

Hypotenuse =AC = q

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 + (√(q² – p²))² = (q)²

⇒ (AB)² + (q² – p²) = q²

⇒ (AB)² = q² – q² + p²)

⇒ (AB)² = p²

⇒ AB =√p²

⇒ AB =±p

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

Now, we have to find sin θ

We know that,

The side opposite to angle θ = AB =p

And Hypotenuse = AC =q

So,

Now, LHS = q sin θ

= q = RHS

Hence Proved