**IB Maths Tutors** say that there are a lot of relationships amongst trigonometric ratios like Sin, Cos, Tan etc. There are many relationships that are true for all values of the variables.

We call these relationships trigonometric identities. Our** IB Tutors** can give you a deep understanding of Trigonometric Identities.

If we make a point P (x,y) on the unit circle making an angle

theta with the origin then

x=Sintheta and y=Costheta

squaring and adding both these values we get

x²+y²=Sin²θ+Cos²θ on a unit circle so **Sin²θ+Cos²θ=1 (i)**

we can also write **Sin²θ=1-Cos²θ** and

**Cos²θ=1-Sin²θ**

these are most widely used identities. If we divide equation (i) by Sin and Cos respectively then we get

**Cosec²θ=1+Cot²θ** and

**Sec²θ=1+Tan²θ**

We can also write these relationships by shifting Cot and Tan and thus we get more Trigonometric identities.

So we see that equation (i) in itself is a creator of nine identities in total

**How to Solve Questions Based on Trigonometric Identities-**

Although every trigonometric identities problem is unique in its own way yet we can apply a few simple steps to solve the majority of them

**►First of all, we must check for the common value in the question**

**► Check if any direct Trigonometric identity is given in the question or not, if yes then we should apply it**

**► Convert every given trigonometric ratio in Sin/Cos**

**►Take L.C.M**

**► Simplify everything to get the required result**

For simplification, we can use algebraic identities like- **(a+b)², (a+b)³, a²-b²,** and **a³+b³** etc.

**Let us now try some questions based on the above tricks**

**Question-1 Prove That**

(cosec θ – cot θ)^{2 }= (1-cos θ)/(1+cos θ)

**Solution:**

**L.H.S**

(Cosecθ-Cotθ)²=(1/Sinθ-Cosθ/Sinθ)²

**Question-2**

**Show that**

**Solution:**

**L.H.S.**cos A/(1+sin A) + (1+sin A)/cos A

**Question- 3**

If acosθ – bsinθ = c, prove that

asinθ + bcosθ = ±√a²+b²-c²

**Solution: **

(acosθ – bsinθ)² + (asinθ + bcosθ)²

= (a²cos²θ + b²sin²θ – 2ab sinθcosθ)+ (a²sin²θ + b²cos²θ + 2ab sinθcosθ)

= (a²cos²θ + b²sin²θ ) + (a²sin²θ + b²cos²θ)

= a²(cos²θ +sin²θ) + b²(cos²θ +sin²θ)

=a²+b²

**Here I am posting a few questions based on Trigonometric Identities **

#### Trigonometric identities Worksheet.pdf

You can read and share my other posts on Trigonometry

#### First post- concept one

#### Second post- concept two

#### Third post -Concept three

**double angle formula, compound angle formula**,

**Sine Rule, Cosine Rule**etc so I will keep on posting on this topic