Question

prove that: sin²∅+ cos²∅=1​

Answer 1

Proof

Let ∠ABC be an acute angle . take a point P on BA and Draw PQ⊥BC

Let ∠PBQ = θ

Then

Sinθ = PQ/BP , Cosθ = BQ/BP , Tanθ=PQ/BQ ,

Cosecθ= BP/PQ, Secθ = BP/BQ and Cot= BQ/PQ                   (i)

In Right Triangle PQB

We have

PQ² + BQ² = BP²            by using Pythagoras Theorem                  (ii)

Now Divide by BP² , We get

PQ²/BP² + BQ²/BP² = BP²/BP²

PQ²/BP² + BQ²/BP² = 1

(PQ/BP)² + (BQ/BP)² = 1

By Using (i)st Equation

(Sinθ)² + (Cosθ)² = 1

Sin²θ+ Cos²θ = 1

Hence Proved

Answer 2

Answer:

Let ABC be a right angled triangle right angled at B.

Let angle C ie angle ACB = theeta.

Now by Pythagoras theorem

AC^2= AB^2+BC^2 ……(1)

We know that sin theeta = opposite side/ hypotenuse.

So sin theeta = AB/AC. Similarly

cos theeta= adjacent side/ hypotenuse

So cos theeta = BC/AC

Now sin^2 theeta + cos^2 theeta

= (AB/AC)^2 + (BC/AC)^2

= AB^2/AC^2 + BC^2/AC^2

= (AB^2+BC^2)/AC^2

= AC^2/AC^2 = 1 ( by using (1) )