# Trigonometric Identities

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Trigonometric identities are important to learn because they allow us to solve equations involving trigonometric functions. These identities are based on the fundamental relationships between the angles and sides of triangles, and they can be used to simplify expressions, to find the exact values of trigonometric functions, and to prove other mathematical results. They are a fundamental tool in mathematics, and they are used in a wide range of applications, from geometry and calculus to engineering and physics.

Let us look at identifying some trigonometric identities.

Given the triangle ABC, the identity we already know is

If we have to rephrase the above equation in terms of sin and cos, how can we do it? We know that sin and cos have ratios with the hypotenuse in the denominator. So, we need to divide each term in the equation with:

**This is true for all A such that 0° ≤ A ≤ 90°. So, this is a trigonometric identity**.

If we have to get an identity in terms of tan we need to divide

Is this equation true for A = 0°? Yes, it is. What about A = 90°? Well, tan A and sec A are not defined for A = 90°. So, (3) is true for all A such that 0° A 90°.

Let us see what we get on dividing (1) by BC2 . We get 2 2 2 2 AB BC BC BC = 2 2 AC BC

i.e., 2 2 AB BC BC BC = 2 AC BC i.e., cot2 A + 1 = cosec2 A (4) Note that cosec A and cot A are not defined for A = 0°. Therefore (4) is true for all A such that 0° < A 90°.

Similarly we can find:

Using these identities, we can express each trigonometric ratio in terms of other trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the values of other trigonometric ratios.

Let us see how we can do this using these identities.

If tan A =

Look above and see which identity we can use. We need to choose an identity which has tan in it.

Since,

The inverse of sec A is nothing but

So cos A =

Now that we have cos A, we need to pick an identity which has cos in it.

So sinA =

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**Example 9 :** Express the ratios cos A, tan A and sec A in terms of sin A.

**Solution :** Since

cos A =

This gives cos A =

Hence, tan A =

**Example 10 :** Prove that sec A (1 – sin A)(sec A + tan A) = 1.

**Solution :**

LHS =

=

**Example 11 :** Prove that

**Solution :**

LHS =

=

=

**Example 12 :** Prove that

**Solution :** Since we will apply the identity involving sec θ and tan θ, let us first convert the LHS (of the identity we need to prove) in terms of sec θ and tan θ by dividing numerator and denominator by cos θ,

LHS =

=

=

=

=

which is the RHS of the identity, we are required to prove.

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## Trigonometric Tools

Some of the tools required to solve trignometric ratio problems

**1)Reciprocal rule:**

sinA =

cosA =

tanA =

**2)Quotient rule:**

tanA =

cotA =

**3)Identities:**

**4)Algebric identities:**

Tips to solve problems using Trignometric ratios and identities:

1)Always start from the complex side no. of terms,no. of operations,trignometric ratio,presence of square root in the question.

- Eg:
1 + cosA 1 − cosA = cosecA + cotA ;

start from

as it has

- express in terms of sin & cos when different terms are seen.

- Eg:
tanA 1 − cotA + cotA 1 − tanA = 1 + sinA CosecA ;

- Combine terms into single fraction.

- Eg:
cosA 1 + sinA + 1 + sinA cosA = 2 SecA ;

- Use Pythagoras theorem

- Eg:
sin 2 A 1 − cosA = 1 + secA secA

- Solve L.H.S and R.H.S seperately and prove they are equal.

- Eg:
cosecA − sinA · secA − cosA = 1 tanA + cotA

- Expand / Factorize / Simplify / cancel common terms if needed using above
**Trigonometric Tools**.