Trigonometry builds on a good knowledge of Arithmetic, Geometry, and Algebra. Throughout history, trigonometry has been applied in areas such as navigation, in measuring the height of a building or a mountain.

- Suppose, some students of a school are visiting a lighthouse which is located on one side of the bank of a river. From the point on the other side of the bank, exactly opposite, if a student is looking at the top of the lighthouse a right-angled triangle is formed. Can the student find out the height of the lighthouse without actually measuring it?
- Suppose, you are a professional photographer and you like to measure particular features within an image. Suppose there is a model rocket that will reach a certain altitude and you want to set up your camera some distance away from the launch pad. Can one find the angle at which you set the camera to get a picture of the model rocket at the maximum altitude?
- One fine day, a girl is in the park and a hot air balloon is flying in the air at a certain height from the ground. The girl suddenly spotted the balloon at a point A and got excited. After some time, the balloon moves in the same horizontal direction to some distance and is at point B. She was curious to know the altitude of the hot air balloon at point B from the ground.

In all the above cases, the distance, as well as the height, can be found out by using some mathematical techniques known as, Trigonometry*.*

### Trigonometric Identities

In mathematics, an identity* *is an equation that is verifiable for each and every value of the variables. Similarly, trigonometric identities are the equations involving trigonometric functions which hold true for every value of the variables selected.

**The six trigonometric ratios **

- sine
- cosine
- tangent
- cosecant
- secant, and
- cotangent.

All these trigonometric ratios are derived by using the sides of the right triangle, such as a base, perpendicular, and hypotenuse.

- sin θ= perpendicular / hypotenuse
- cos θ= base / hypotenuse
- tan θ= perpendicular / base
- cosec θ= hypotenuse / perpendicular
- sec θ= hypotenuse / base
- cot θ= base / perpendicular

**Mnemonics for the above relations **

Some (sin θ) people (perpendicular) had (hypotenuse) curly (cos θ) black (base) hair (hypotenuse) turned(tan θ) permanently (perpendicular) brown (base).

Lets learn about each type of trigonometric identities in detail, Observe from the above relations:

- sin θ = 1/cosecθ or, cosec θ = 1/sinθ
- cos θ = 1/secθ or, sec θ = 1/cosθ
- tan θ = 1/cotθ or, cot θ = 1/tanθ

These above identities are known as** Reciprocal Trigonometric Identities** and

- tan θ = sin θ /cos θ
- cot θ = cos θ /sin θ

These types of identities are known as **Ratio Trigonometric Identities.**

**Pythagorean Trigonometric Identities**

From the Pythagoras theorem that has learned in earlier classes, In a right-angled triangle,

**Perpendicular ^{2 }+ Base^{2 }= Hypotenuse^{2}**

Now dividing both sides by Hypotenuse^{2} on both the sides,

**Perpendicular ^{2} / Hypotenuse^{2} + Base^{2} / Hypotenuse^{2} = Hypotenuse^{2} / Hypotenuse^{2}**

As it is known,

sin θ = perpendicular/hypotenuse;

cos θ = base /hypotenuse

Therefore,

- sin
^{2}θ + cos^{2}θ = 1

The other two Pythagorean trigonometric identities are which can be derived in the same way:

- 1 + tan
^{2 }θ = sec^{2}θ - 1 + cot
^{2}θ = cosec^{2}θ

Another type of identity is there which is known as Complementary and Supplementary Trigonometric Identities. Let’s learn about it,

**Complementary and Supplementary Trigonometric Identities**

From the definition of the complementary angle, we know that when the sum of two angles is equal to 90° then that pair of angles is known as the complementary angle. Therefore,

- sin (90°- θ) = cos θ
- cos (90°- θ) =sin θ
- cosec (90°- θ) = sec θ
- sec (90°- θ) = cosec θ
- tan (90°- θ) = cot θ
- cot (90°- θ) = tan θ

Similarly, when the sum of two angles is equal to 180° then that pair of angles is known as supplementary angles. So the supplementary identities are as follows:

- sin (180°- θ) = sinθ
- cos (180°- θ) = -cos θ
- cosec (180°- θ) = cosec θ
- sec (180°- θ) = -sec θ
- tan (180°- θ) = -tan θ
- cot (180°- θ) = -cot θ

**Sum and Difference Trigonometric Identities** constitute a significant part of the trigonometric identity.

- sin (A+B) = sin A cos B + cos A sin B
- sin (A-B) = sin A cos B – cos A sin B
- cos (A+B) = cos A cos B – sin A sin B
- cos (A-B) = cos A cos B + sin A sin B
- tan (A+B) = (tan A + tan B)/(1 – tan A tan B)
- tan (A-B) = (tan A – tan B)/(1 + tan A tan B)

**Double,** **Half**,** and Triple Angles Trigonometric Identities**

Double angle formulas ⇢ From the above sum and difference formulas:

sin (A+B) = sin A cos B + cos A sin B

By replacing, A = B = θ,

sin (θ + θ) = sinθ cosθ + cosθ sinθ

- sin 2θ = 2sinθcosθ

The other double angle formulas are:

- cos 2θ = cos
^{2}θ – sin^{2 }θ

= 2cos^{2} θ – 1

= 1 – 2sin^{2 }θ

- tan 2θ = (2tanθ)/(1 – tan
^{2}θ )

*Half Angle Formulas ⇢ *From the above double angle formulas we have got,

cos 2θ = 1 – 2 sin^{2} θ

or, 2 sin^{2} θ = 1- cos 2θ

or, sin^{2 }θ = (1 – cos2θ)/(2)

or, sinθ = ±√[(1 – cos 2θ)/2]

By substituting θ by θ/2 on both sides,

- sin (θ/2) = ±√[(1 – cos θ)/2]

The other half-angle formulas are:

- cos (θ/2) = ±√(1 + cosθ)/2
- tan (θ/2) = ±√[(1 – cosθ)(1 + cosθ)]

Triple angle Formulas ⇢ Again from the sum and difference formulas, we can derive the triple angle formulas,

- sin3θ = sin(2θ + θ)

= sin2θcosθ + cos2θsinθ

= (2sinθcosθ)cosθ + (1 – 2sin2θ)sinθ

= 2sinθcos2θ + sinθ – 2sin3θ

= 2sinθ(1 – sin2θ) + sinθ – 2sin3θ

= 2sinθ – 2sin3θ + sinθ – 2sin3θ

*= 3sinθ – 4 sin ^{3}θ*

The other triple angle formulas are:

- cos3θ = 4cos
^{3}θ – 3cosθ - sin3θ = 3sinθ – 4 sin
^{3}θ - tan3θ = 3tanθ – tan
^{3}θ/1- 3tan^{2 }θ

There are a few other identities that are not derived from right-angled. One of them is** Sine and Cosine Rule Trigonometric Identities. **For a triangle with sides ‘a’, ‘b’, and ‘c’ and the respective opposite angles of the triangle are A, B, and C, sine rule can be given as:

- a/sin A = b/sin B = c/sin C
- sin A/a = sin B/b = sin C/c
- a/b = sin A/sin B
- a/c = sin A/sin C
- b/c = sin B/sin C

Cosine rule for a triangle with sides ‘a’, ‘b’, and ‘c’ and the respective opposite angles are A, B, and C can be given as,

- a
^{2}= b^{2}+ c^{2}– 2bc × cosA - b
^{2}= c^{2}+ a^{2}– 2ca × cosB - c
^{2}= a^{2}+ b^{2}– 2ab × cosC

### Prove that (sec A + tan A)(1 – sin A) = cos A

**Solution:**

From the reciprocal identities of trigonometry, we know that sec A = 1/cos A;

From the rule of ratio identities, tan A = sin A/cos A

Therefore,

(sec A + tan A) (1 – sin A)

= (1/cos A + sin A/cos A)(1−sin A)

= {(1+sin A )/cos A}(1 − sin A)

= (1− sin A)(1 + sin A)/cos A

(a + b)(a – b)= a

^{2}– b^{2,}So,(1 + sin A)(1 – sin A) = 1 – sin

^{2}A.= (1 – sin

^{2 }A)/cos Asin

^{2}A + cos^{2}A = 1. So,1 – sin

^{2}A = cos^{2}APutting it in above expression,

= cos

^{2}A/cos A= cos A

Hence, the value of (sec A + tan A) (1 – sin A) is equal to cos A.

### Similar Problems

**Question 1: Prove (1 – cos A)(1 + cos A)(1 + cot ^{2}A) = 1**

**Solution:**

L.H.S(1 – cos A)(1 + cos A)(1 + cot

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

^{2}A)(a + b)(a – b)= a

^{2}– b^{2}, So,(1 + cos A)(1 – cos A) = 1 – cos

^{2}A.= (1 – cos

^{2}A)(1 + cot^{2}A)sin

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

^{2}A = 1 – cos^{2}ATherefore,

= sin

^{2}A × (1 + cot^{2}A)= sin

^{2}A + sin^{2}A × cot^{2}AFrom the rule of ratio identities, cot A = cos A/sin A ,

= sin

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

^{2}A + cos^{2}A= 1

= R.H.S

**Question 2: Proof cos A/(1 + sin A) = (1 – sin A)/cos A**

**Solution:**

L.H.Scos A/(1 + sin A)

Multiplying both numerator and denominator by (1 – sin A)

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

= (cos A)(1 – sin A)/(1 – sin

^{2}A)= (cos A)(1 – sin A )/ cos

^{2}A= (1 – sin A)/ cos A

R.H.S

**Question 3: Prove, tan θ sin θ + cos θ = sec θ**

**Solution:**

L.H.Stan θ sin θ + cos θ

From the rule of ratio identities we know that , tan θ=sin θ /cos θ

= tan θ sin θ + cos θ

= (sin θ/cos θ) ⋅ sin θ + cos θ

= (sin

^{2}θ/cos θ) + cos θ= (sin

^{2}θ/cos θ) + (cos^{2}θ/cosθ)= (sin

^{2}θ + cos^{2}θ) / cos θsin

^{2}θ + cos^{2}θ = 1,= 1 / cos θ

From the reciprocal identities of trigonometry we know that, sec θ =1/cos θ

= sec θ

= R.H.S

**Question 4: Prove cos A cosec A tan A = 1 **

**Solution:**

From the reciprocal identities of trigonometry, cosec A = 1/sin A

From the rule of ratio identities, tan A = sin A /cos A

L.H.Scos A cosec A tan A

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

= (cos A/sin A) ( sin A/cos A )

= cos A sin A/sin A cos A

= 1

= R.H.S