# How to prove that sec²θ + csc²θ = sec²θ × csc²θ?

Trigonometry is a study of the properties of triangles and trigonometric functions. It is used a lot in engineering, science, for building video games, and more. It deals with the relationship between ratios of the sides of a right-angled triangle with its angles. These ratios which are used for the study of this relationship are called **Trigonometric ratios**.

### Trigonometric Ratios

There are six basic trigonometric ratios that establish the co-relation between sides of a right triangle with the angle. If θ is the angle formed between the base and hypotenuse, in a right-angled triangle (as shown in the above figure), then

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sin(θ) = Perpendicular/Hypotenuse

cos(θ) = Base/Hypotenuse

tan(θ) = Perpendicular/Base

The values of the other three function ,that is, cosec(θ), sec(θ), cot(θ) depend on sin(θ), cos(θ), tan(θ) respectively.

cot (θ) = 1/tan (θ) = Base/Perpendicular

sec (θ) = 1/cos (θ) = Hypotenuse/Base

cosec (θ) = 1/sin (θ) = Hypotenuse/Perpendicular

### Trigonometric Identities

A trigonometric equation that holds for every viable value for an input variable on which it is defined is called trigonometric identities. One identity that we are familiar with is the Pythagorean Identity,

#### Sin^{2}θ + Cos^{2}θ = 1

Divide both sides of the Pythagorean Identity by cosine squared, which is allowed, then,

[cos^{2}θ + sin^{2}θ]/cos^{2}θ = 1/cos^{2}θ

cos^{2}θ/cos^{2}θ + sin^{2}θ/cos^{2}θ = 1/cos^{2}θ

1 + sin^{2}θ/cos^{2}θ = 1/cos^{2}θ

(using the definition tanθ = sinθ/cosθ)

1 + tan^{2}θ = 1/cos^{2}θ

(using the definition secθ = 1/cosθ)

1 + tan^{2}θ = sec^{2}θ

Therefore the next trigonometric identity is,

**1 + tan ^{2}θ = sec^{2}θ**

Similarly, if divide both sides of the Pythagorean Identity by sine squared then we obtained the last identity,

**cot ^{2}θ + 1 = cosec^{2}θ**

### How to prove that sec^{2}θ + csc^{2}θ = sec^{2}θ × csc^{2}θ?

**Proof:**

To solve above problem we require below specified trigonometric identities and ratios :

sec(θ) = 1/cos(θ) and cosec(θ) = 1/sin(θ) ⇢ Eq. 1

sin^{2}θ + cos^{2}θ =1 ⇢ Eq. 2

sec^{2}θ = 1 + tan^{2}θ ⇢ Eq. 3

cosec^{2}θ = 1+ cot^{2}θ ⇢ Eq. 4

There are **two** ways to solve this problem

**1. To prove LHS = RHS using identities**

LHS = sec

^{2}(θ) + cosec^{2}(θ)=(1+ tan

^{2}θ) + (1+cosec^{2}θ) (from 3 and 4)=2 + tan

^{2}θ + cosec^{2}θRHS= sec

^{2}θ × cosec^{2}θ=(1+tan

^{2}θ) × (1+cot^{2}θ)=2 + tan

^{2}θ + cot^{2}θTherefore, LHS = RHS.

**2. By using trigonometric ratios**

LHS= sec

^{2}θ + cosec^{2}θ(from 1), [1/cos

^{2}θ ] + [1/sin^{2}θ]= [sin

^{2}θ + cos^{2}θ] / [cos^{2}θ × sin^{2}θ](from 2), 1/[cos

^{2}θ × sin^{2}θ]= sec

^{2}θ × cosec^{2}θ = RHS

Hence, the given trigonometric equation can be solved in two ways as mentioned above

Therefore, sec^{2}(θ) + cosec^{2}(θ) = sec^{2}(θ) × cosec^{2}(θ).

### Similar Problems

**Question 1: Prove, tan ^{4}(θ) + tan^{2}(θ) = sec^{4}(θ) – sec^{2} (θ) [Hint: take tan^{2}(θ) as common]**

**Solution:**

LHS= tan

^{4}θ + tan^{2}θ = tan^{2}θ (tan^{2}θ + 1)= (sec

^{2}θ – 1) (tan^{2}θ + 1) {since, tan^{2}θ = sec^{2}θ – 1}=(sec

^{2}θ – 1) sec^{2}θ {since, tan^{2}θ + 1 = sec^{2}θ}=sec

^{4}(θ) – sec^{2}(θ) = RHSHence proved.

**Question 2: Prove, cos θ / [(1 – tan θ)] + sin θ / [(1 – cot θ)] = sin θ + cos θ**

**Solution:**

LHS = cos θ / [(1 – tan θ)] + sin θ / [(1 – cot θ)]

=cos θ / [1 – (sin θ/cos θ)] + sin θ/[1 – (cos θ/sin θ)]

= cos θ / [(cos θ – sin θ/cos θ] + sin θ / [(sin θ – cos θ/sin θ)]

= cos

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

^{2}θ – sin^{2}θ)/(cos θ – sin θ)= [(cos θ + sin θ)(cos θ – sin θ)] / (cos θ – sin θ)

= (cos θ + sin θ) = RHS

Hence proved.

**Question 3: Prove, (tan θ + sec θ – 1)/(tan θ – sec θ + 1) = (1 + sin θ) × sec θ**

**Solution:**

LHS = (tan θ + sec θ – 1)/(tan θ – sec θ + 1)

= [(tan θ + sec θ) – (sec

^{2}θ – tan^{2}θ)]/(tan θ – sec θ + 1) [ sec^{2}θ – tan^{2}θ = 1]= [ (tan θ + sec θ) – (sec θ + tan θ) (sec θ – tan θ)] / (tan θ – sec θ + 1)

= [ (tan θ + sec θ) × (1 – sec θ + tan θ)] / (tan θ – sec θ + 1)

= [(tan θ + sec θ) × (tan θ – sec θ + 1)] / (tan θ – sec θ + 1)

= (tan θ + sec θ)

= (sin θ/cos θ) + (1/cos θ)

= (sin θ + 1) / cos θ

= (1 + sin θ) × secθ = RHS ; Hence proved.

**Question 4:** = **cosec θ – cot θ [Hint: Multiply numerator and denominator by (sec θ – 1)]**

**Solution:**

LHS =

= (multiply numerator and denominator by (sec θ – 1))

=

=√[(sec θ -1)

^{2 }/ tan^{2}θ ] {sec^{2}θ = 1 + tan^{2}θ ⇢ sec^{2}θ – 1 = tan^{2}θ}= (sec θ – 1) / tan θ

= (sec θ/tan θ) – (1/tan θ)

= [(1/cos θ) / (sin θ/cos θ)] – cot θ

= [(1/cos θ) × (cos θ/sin θ)] – cot θ

= (1/sin θ) – cot θ

= cosec θ – cot θ = RHS, Hence proved.

**Question 5: Prove, (sin θ+cosec θ) ^{2}+(cos θ+sec θ)^{2}=7+tan^{2}(θ)+cot^{2}(θ)**

**Solution:**

LHS = {sin

^{2}θ + cosec^{2}θ + 2 sinθ cosecθ } + {cos^{2}θ+ sec^{2}θ +2 cosθ secθ}={sin

^{2}θ + cosec^{2}θ + 2} + { cos^{2}θ + sec^{2}θ + 2}=sin

^{2}θ + cos^{2}θ + sec^{2}θ + cosec^{2}θ + 4=sec

^{2}θ + cosec^{2}θ + 5 [sin^{2}θ +cos^{2}θ = 1]= (1+ tan

^{2}θ) + (1+ cot^{2}θ) +5 [sec^{2}θ = 1 + tan^{2}θ ; cosec^{2}θ = 1+ cot^{2}θ ]=7+ tan

^{2}(θ) + cot^{2}(θ) = RHSHence proved.