Trigonometry is a branch of mathematics which deals with the real life problems related to angles. 'Trigo' means three sides. So, trigonometry is basically a branch of mathematics which relates to the study of three sided figures i.e. Triangles. Trigonometry is a branch of Math studying the relationship between angles and lengths of the sides. In trigonometry, we take into consideration, a Right Angle triangle with acute angles. These Relations are used to solve real life problems related to inclination at a particular point. Trigonometry is used to find the angles or the distance between the two points, when the angle of inclination is known. Trigonometric functions such as sine, cosine and tangent are used in computations in Trigonometry. Trigonometry is an important branch of mathematics which deals with ratios and relationships between angles and sides of triangles, especially right angled triangles. The study of trigonometry mainly revolves around the three functions. The values of sine, cosine and tangent vary with the change of angles. Example, sin (0) = 0, sin (30) = 1/2. |

**Trigonometric Pythagorean Identities:**

^{2 }a + cos

^{2}a = 1

^{2 }a = sec

^{2}a

^{2}a = csc

^{2}a

**Trigonometric Identities for Reciprocal:**

csc a = $\frac{1}{\sin a}$

sec a = $\frac{1}{\cos a}$

cot a = $\frac{1}{\tan a}$

sin a = $\frac{1}{\csc a}$

cos a = $\frac{1}{\sec a}$

tan a = $\frac{1}{\cot a}$

**Trigonometric Identities for Quotient:**

tan a = $\frac{\sin a}{\cos a}$

cot a = $\frac{\cos a}{\sin a}$

**Trigonometric Identities for Even - Odd:**

cos (-a) = cos a

sin (-a) = - sin a

tan (-a) = - tan a

sec (-a) = sec a

csc (-a) = - csc a

cot (-a) = - cot a

**Trigonometric Identities for double Angle Formula:**

sin (2a) = 2 sin a cos a

cos (2a) = 1 - 2 sin

^{2 }a

tan (2a) = $\frac{2 \tan a}{1 - \tan ^{2} a}$

Now, lets try proving few trigonometric identities.

We know that,

sin $\theta$ = $\frac{\text{opposite}}{\text{hypotenuse}}$

cos $\theta$ = $\frac{\text{adjacent}}{\text{hypotenuse}}$

tan $\theta$ = $\frac{\text{opposite}}{\text{adjacent}}$

According to Pythagorean theorem

a

^{2}+ b

^{2}= c

^{2}

By dividing c

^{2}both the side, we get

$\frac{a^{2}}{c^{2}}$ + $\frac{b^{2}}{c^{2}}$ = $\frac{c^{2}}{c^{2}}$

i.e. ${(\frac{a}{c})^2}$ + ${(\frac{a}{c})^2}$ = 1

**cos**.

^{2}$\theta$ + sin^{2}$\theta$ = 1Now, divide both the sides by cos

^{2}$\theta$

$\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$ + $\frac{\cos ^{2} \theta}{\cos ^{2} \theta}$ = $\frac{1}{\cos ^{2} \theta}$

**tan**Hence proved.

^{2}$\theta$ + 1 = sec^{2}$\theta$Given below are few problems based on Trigonometry Identities:

### Solved Examples

**Question 1:**Prove (cos A - sin A +1) / (Cos A + sin A - 1) = cosec A + cot A

**Solution:**

$\frac{\cos A - \sin A + 1}{\sin A}$

$\frac{\cos A + \sin A - 1}{\sin A}$

We get $\frac{\cot A - 1 + \csc A}{\cot A + 1 - \csc}$

= $\frac{\cot A - (1 - \csc A)}{\cot A + (1 - \csc A)}$

Now, let us multiply the numerator and denominator by cot A - (1 - cosec A)

= $\frac{(\cot A - (1 - \csc A))^2}{\cot^2 A - (1 - \csc A)^2}$

= $\frac{\cot^2 A + 1 + \csc^2 A - 2\csc A - 2\cot A + 2\cot A \csc A}{2\csc A - 2}$

= $\frac{2 \csc\ A(\csc A - 1) + 2 \cot A(\csc A - 1)}{2 \csc A - 2}$

= $\frac{2(\csc A + \cot A)(\csc A - 1)}{2(\csc A - 1)}$

By cancelling the common terms, we get

(cos A - sin A + 1)/ (Cos A + sin A - 1) = cosec A + cot A.

Hence proved.

**Question 2:**Solve (sec A + tan A)(1 - sin A)

**Solution:**

= $(\frac{1}{\cos A})$ + $(\frac{\sin A}{\cos A})$ - sin A$(\frac{1}{\cos A})$ - sin A$(\frac{\sin A}{\cos A})$

= $\frac{1 - \sin^2 A}{\cos A}$

= $\frac{\cos^2 A}{\cos A}$

= cos A

Hence proved.