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

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