Question

prove that sin^2q + cos^2q = 1 for an acute angle q of a right angled triangle

Answer 1

Answer:-

Given:

q is an acute angle i.e., 0 < q < 90° in a Right angled triangle.

Let us assume that the ∆le is ∆ABC , right angled at B and q is at A.

Let the adjacent side of q be x (AB = x) and it's opposite side be y(BC = y).

Using Pythagoras Theorem,

→ (Hypotenuse)² = x² + y²

→ (AC)² = x² + y²

→ (AC) = √x² + y²

We know that,

sin A = Opposite side/Hypotenuse

Hence,

sin q = y/√x² + y²

Similarly,

Cos q = Adjacent side/Hypotenuse = x/√x² + y²

We have to prove that:

sin² q + cos² q = 1

Putting the values of sin q and Cos q we get,

→ (y/√x² + y²)² + (x/√x² + y)²) = 1

→ y²/x² + y² + x²/x² + y² = 1

→ (x² + y²)/(x² + y²) = 1

→ 1 = 1

→ LHS = RHS

Hence, we can say that sin² q + cos² q = 1.

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