Trigonometry being a branch of mathematics, which deals with the ratios of right angled triangles. Trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables.

# Proving trigonometric identities using Pythagoras

The above triangle is a right angled triangle. Let us prove **trigonometric identities** by using Pythagoras theorem.

Pythagoras theorem states that AC^{2} = AB^{2} + BC^{2}.

Therefore, dividing each term of the theorem by AC^{2}, we get:

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

i.e. (AB/AC)^{2} + (BC / AC)^{2} = (AC / AC)^{2}.

i.e. (cos A)^{2} + (sin A)^{2} = 1

Therefore, **cos ^{2} A + sin^{2} A = 1**

This is true for all A such that 0^{0} ≤ A ≤ 90^{0}. Therefore, this is a trigonometric identity.

Now, dividing each term by (AB)^{2}, we get :

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

i.e. (AB/AB)^{2} + (BC / AB)^{2} = (AC / AB)^{2}.

Therefore, **1 + tan ^{2} A = sec^{2} A**

This is also true for all A such that 0^{0} ≤ A ≤ 90^{0}. Therefore, this is also a trigonometric identity.

Now, dividing each term by (BC)^{2}, we get :

i.e. (AB/BC)^{2} + (BC / BC)^{2} = (AC / BC)^{2}.

Therefore, **cot ^{2} A + 1 = cosec^{2} A**

This is also a trigonometric identity.

Therefore there are 3 trigonometric identities. They are

**cos**^{2}A + sin^{2}A = 1**1 + tan**^{2}A = sec^{2}A**cot**^{2}A + 1 = cosec^{2}A

## Problems on trigonometric identities

** **Let us see some problems on these identities. The problems will be based on the above three trigonometric identities. Also knowledge on trigonometric ratios helps in solving these problems.

**1) ****Solve (sec A + tan A) (1 – sin A)**

Solution: (sec A + tan A) (1 – sin A) = sec A + tan A – sin A sec A – sin A tan A

= (1/cos A) + (sin A/cos A) – sin A (1/cos A) – sin A (sin A/cos A)

= (1 + sin A – sin A – sin^{2} A)/ cos A

= (1- sin^{2} A)/ cos A [we know that **cos ^{2} A + sin^{2} A = 1]**

** **= cos^{2} A/ cos A** **

= **cos A**

**2) ****Prove (1+ sec A)/ sec A = sin ^{2}A / (1 – cos A)**

Solution: Let us take the RHS of the equation

**sin ^{2}A / (1 – cos A) **= (1 – cos

^{2}A)/(1 – cos A) [we know that

**cos**

^{2}A + sin^{2}A = 1]** **= (1 + cos A)(1 – cos A) / (1 – cos A)

= (1 + cos A)

Multiply and divide by sec A

= (1 + cos A) sec A/sec A

= (**sec A + 1) / sec A**

** Hence it is proved.**

**3) ****Prove (cos A – sin A + 1)/ (cos A + sin A – 1) = cosec A + cot A.**

Solution: Let us take the left hand side of the equation. Divide numerator and denominator by sin A.

(cos A – sin A + 1)/ (cos A + sin A – 1) = (cot A – 1 + cosec A)/ (cot A + 1 – cosec A)

= [cot A – (1 – cosec A)]/[cot A + (1-cosec A)]

** Multiply numerator and denominator by cot A – (1 – cosec A)**

= [cot A – (1 – cosec A)]^{2}/ [cot^{2}A – (1-cosec A)^{2}]

= [cot^{2}A + (1- cosec A)^{2} – 2 cot A(1- cosec A)]/ [cot^{2} A – 1 – cosec^{2} A + 2 cosec A ]

= [**cot ^{2}A + 1** + cosec

^{2}A - 2 cosec A – 2 cot A + 2 cot A cosec A]/[2 cosec A – 2]

= [cosec^{2}A + cosec^{2} A - 2 cosec A – 2 cot A + 2 cot A cosec A]/[2 cosec A – 2]

= [2cosec^{2} A - 2 cosec A – 2 cot A + 2 cot A cosec A]/[2 cosec A – 2]

= [2 cosec A(cosec A – 1) + 2 cot A(cosec A – 1)]/[2 cosec A – 2]

= (2 cosec A + 2 cot A)(cosec A – 1)/2(cosec A – 1)

= 2(cosec A + cot A)/2

= **cosec A + cot A**

Hence it is proved.

### Practice problems on trigonometric identities

Solve the problems given below using trigonometric identities. Answers are also provided for these problems. The practice problems can be solved, once you understood the example problems given above.

1) Solve 1/(cot^{2}B + 1) + 1/(tan^{2}B + 1)

2) Find tan^{2}Bcos^{2}B + 1 – sin^{2}B

3) Solve (tanB sinB) / (sec^{2}B – 1)

**Answers:**

1) 1

2) 1

3) Cos B