We use different identity in trigonometry to solve the problems. We may prove this identity with the help of **Pythagoras Theorem.**

# Pythagoras Theorem

Pythagoras theorem says-

In a right angle triangle square of the larger side(hypotenuse) is equal to the sum of the squares of the other two sides(Perpendicular & Base).

**H² = P² + B²**

Where, H = Hypotenuse, P = Perpendicular

*(opposite to angle θ)*& B = Base.

**1.) How to prove sin²θ+cos²θ = 1 with the help of Pythagoras Theorem?**

Solve-

We know that H² = P² + B²

And according to the above picture we have-

AC² = BC² + AB²

To prove the identity we will divide the above expression by AC²

*(Because we want to make 1 to AC in the above expression)*AC²/AC² = BC²/AC² + AB²/AC²

So as per figure & trigonometry, we will get-

**1 = Sin²θ + Cos²θ**{BC/AC = Sinθ, AB/AC = Cosθ}

Hence Proved

## 2.) How to prove 1 + tan²θ = sec²θ with the help of Pythagoras Theorem?

We know that H² = P² + B²

And according to the above picture we have-

And according to the above picture we have-

AC² = BC² + AB²

To prove the identity we will divide the above expression by AB²

*(Because we want to make 1 to AB in the above expression)*

*AC²/AB² = BC²/AB² + AB²/AB²*

So as per figure & trigonometry, we will get-

**sec²θ =**

**tan²θ + 1**

**{AC/AB = secθ, BC/AB = tanθ}**

Or

**tan²θ + 1 = sec²θ**

**Hence Proved**

## 3.) How to prove cot²θ + 1 = cosec²θ with the help of Pythagoras Theorem?

We know that H² = P² + B²

And according to the above picture we have-

AC² = BC² + AB²

To prove the identity we will divide the above expression by BC²

*(Because we want to make 1 to BC in the above expression)*

*AC²/BC² = BC²/BC² + AB²/BC²*

So as per figure & trigonometry, we will get-

**cosec²θ =**

**1 + cot²θ**

**{AC/BC = cosecθ, AB/BC = cotθ}**

Or

**1 + cot²θ = cosec²θ**

**Hence Proved**