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

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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.

right-angled triangle

Trigonometric Ratios

There are six basic trigonometric ratios that establish the correlation 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 

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,

Sin2θ + Cos2θ = 1 

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

[cos2θ + sin2θ]/cos2θ = 1/cos2θ

cos2θ/cos2θ + sin2θ/cos2θ = 1/cos2θ

1 + sin2θ/cos2θ = 1/cos2θ

(using the definition tanθ = sinθ/cosθ)

1 + tan2θ = 1/cos2θ

(using the definition secθ = 1/cosθ)

1 + tan2θ = sec2θ

Therefore the next trigonometric identity is,

1 + tan2θ  = sec2θ

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

cot2θ + 1 =  cosec2θ

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

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

sin2θ + cos2θ =1  ⇢ Eq. 2

sec2θ = 1 + tan2θ  ⇢ Eq. 3

cosec2θ  = 1+ cot2θ  ⇢ Eq. 4

There are two ways to solve this problem

1. To prove LHS = RHS using identities

LHS = sec2(θ) + cosec2(θ)

=(1+ tan2θ) + (1+cosec2θ)   (from 3 and 4)

=2 + tan2θ + cosec2θ

RHS= sec2θ × cosec2θ

=(1+tan2θ) × (1+cot2θ)

=2 + tan2θ + cot2θ

Therefore, LHS = RHS.

2. By using trigonometric ratios

LHS= sec2θ + cosec2θ

(from 1), [1/cos2θ ]  +  [1/sin2θ]

=  [sin2θ + cos2θ] / [cos2θ × sin2θ]

(from 2), 1/[cos2θ × sin2θ]

= sec2θ × cosec2θ = RHS

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

Therefore, sec2(θ) + cosec2(θ) = sec2(θ) × cosec2(θ).

Similar Problems

Question 1: Prove, tan4(θ) + tan2(θ) = sec4(θ) – sec2 (θ)   [Hint: take tan2(θ) as common]

Solution:

LHS= tan4θ + tan2θ =  tan2θ (tan2θ + 1)

= (sec2θ – 1) (tan2θ + 1)        {since, tan2θ = sec2θ – 1}

=(sec2θ – 1) sec2θ           {since, tan2θ + 1 = sec2θ}

=sec4(θ) – sec2(θ)  =  RHS

Hence 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)]

= cos2θ/(cos θ – sin θ) + sin2θ/(cos θ – sin θ)

= (cos2θ – sin2θ)/(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) – (sec2θ – tan2θ)]/(tan θ – sec θ + 1)       [ sec2θ – tan2θ = 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: \sqrt\frac{sec\theta-1}{sec\theta+1}    = cosec θ – cot θ     [Hint: Multiply numerator and denominator by (sec θ – 1)]

Solution:

LHS =  \sqrt\frac{sec\theta-1}{sec\theta+1}

\sqrt\frac{(sec\theta-1)(sec\theta-1)}{(sec\theta-1)(sec\theta+1)}           (multiply numerator and denominator by (sec θ – 1))

\sqrt\frac{(sec\theta-1)^2}{(sec^2\theta-1)}

=√[(sec θ -1)2 / tan2θ ]           {sec2θ = 1 + tan2θ ⇢ sec2θ – 1 = tan2θ}

= (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+tan2(θ)+cot2(θ)

Solution:

LHS = {sin2θ + cosec2θ + 2 sinθ cosecθ } + {cos2θ+ sec2θ +2 cosθ secθ}

={sin2θ + cosec2θ + 2} + { cos2θ + sec2θ + 2}

=sin2θ + cos2θ + sec2θ + cosec2θ + 4

=sec2θ + cosec2θ + 5                            [sin2θ +cos2θ = 1]

= (1+ tan2θ) +  (1+ cot2θ) +5              [sec2θ = 1 + tan2θ ;  cosec2θ = 1+ cot2θ ]

=7+ tan2(θ) + cot2(θ) = RHS 

Hence proved.



Last Updated : 26 Dec, 2023
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