# Simplify the expression cos(pi/2 – x) sec x

Trigonometry is defined as a branch of mathematics that defines the relationships between sides and angles of triangles. There are basically six angles and their function for calculation in trigonometry. All angles have fixed values and they can also be defined by the ratio of the lengths of the sides in a right-angled triangle. For instance, cos60° is equal to 1/2 and cos30° is equal to √3/2, cos∅ (∅ is the angle) is also defined as the ratio of base to its hypotenuse in a right-angled triangle, and so on. The six trigonometric ratios are:

- sine (sin)
- cosine (cos)
- tangent (tan)
- cotangent (cot)
- cosecant (cosec)
- secant (sec)

### Trigonometric Identities

Trigonometric Identities are the equalities that involve trigonometry functions and hold true for all the values of variables given in the equation. There are various distinct trigonometric identities involving the side length as well as the angle of a triangle. The trigonometric identities are applied only for a right-angle triangle. All the trigonometric identities are based on the six trigonometric ratio.

### Basic trigonometric identities

There are trigonometric identities derived that are used in the questions and calculations. They are reciprocal identities, identities of opposite angles, identities of complementary angles, identities of supplementary angles, product-sum identities, identities of product. Let’s take a look at these identities.

**Reciprocal Identities**

- sin θ = 1/cosec θ or cosec θ = 1/sin θ
- cos θ = 1/sec θ or sec θ = 1/cos θ
- tan θ = 1/cot θ or cot θ = 1/tan θ

There are three Pythagorean trigonometric identities that are based on the Pythagoras theorem or right-triangle theorem.

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

**Identities of Opposite Angles**

The opposite angle trigonometric identities are as:

- sin (-θ) = – sin θ
- cos (-θ) = cos θ
- tan (-θ) = – tan θ
- cot (-θ) = – cot θ
- sec (-θ) = sec θ
- cosec (-θ) = -cosec θ

**Identities of Complementary Angles**

Complementary angles are those two angles that have a sum equal to 90°. All complementary angles are positive because they all are lying in the 1st quadrant.

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

**Identities of Supplementary Angles**

Supplementary angles are those two angles have sum equal to 180°. different trigonometric angles are shown as:

- 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

**Product-Sum Identities**

The product-sum trigonometric identities are as:

- sin x + sin y = 2 sin(x + y)/2 . cos(x – y)/2
- cos x + cos y = 2 cos(x + y)/2 . cos(x – y)/2
- sin x – sin y = 2 cos(x + y)/2 . sin(x – y)/2
- cos x – cos y = -2 sin(x + y)/2 . sin(x – y)/2

**Identities of Products**

These identities are as:

- sin x . sin y = [cos(x – y) – cos (x + y)]/2
- sin x . cos y = [Sin (x + y) – Sin (x – y)]/2
- cos x . cos y = [Cos (x + y) – Cos (x – y)]/2

### Simplify the expression cos(π/2 – x) sec x

**Solution****:**

In the given expression cos(π/{2} – x) sec x

It is known cos(π/2 – x) = sin x

Now, put sin x into the expression,

= sin x . sec x = sin x . 1/cos x ⇢ (sec x = 1/cos x)

= sin x/ cos x =

tan x

### Sample problem

**Question 1: Prove: (1 – sin x)/(1 + sin x) = (sec x – tan x) ^{2}**

**Solution:**

L.H.S = (1 – sin x)/(1 + sin x)

= (1 – sin x)

^{2}/(1 – sin x) (1 + sin x) ⇢ [Multiply both numerator and denominator by (1 – sin x)= = (1 – sin x)

^{2}/(1 – sin^{2}x)= (1 – sin x)

^{2}/(cos^{2}x), [Since, sin^{2}θ + cos^{2}θ = 1 ⇒ cos^{2}θ = 1 – sin^{2 }θ]= {(1 – sin x)/cos x}

^{2}= (1/cos x – sin x/cos x)

^{2}= (sec x – tan x)

^{2}= R.H.S. hence proved.

**Question 2: In the expression Find the value of x: cos x = 2 sin 45° cos 45° – sin 30°.**

**Solution:**

cos x = 2 (1/√2). (1/√2) – 1/2

cos x = 2 (1/2) – 1/2

cos x = 1 – 1/2

cos x = 1/2

x = cos

^{-1}(1/2)x = 60°

**Question 3: Show that (1 – cos ^{2 }A) cosec^{2} A = 1.**

**Solution:**

we know , 1 – cos

^{2}A = sin^{2}A= sin

^{2}A . cosec^{2}A = sin^{2}A / sin^{2}A =1 hence proved.

**Question 4: Show that, tan θ sin θ + cos θ = sec θ**

**Solution:**

L.H.S = 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^{2 }θ)/ cos θ= 1/cos θ = sec θ

= R.H.S = Hence proved.

**Question 5: Write the relation between **

**sin and cos****sec and cos****tan with sin and cos.**

**Solution:**

- sin
^{2 }x + cos^{2}x = 1- sec x = 1/cos x
- tan x = sin x / cos x

**Question 6: Prove: tan ^{4}θ + tan^{2}θ = sec^{4}θ – sec^{2}θ**

**Solution:**

L.H.S = tan

^{4}θ + tan^{2}θ= tan

^{2}θ (tan^{2}θ + 1)tan

^{2}θ = sec^{2}θ – 1tan

^{2}θ + 1 = sec^{2}θThen,

= (sec

^{2}θ – 1)(sec^{2}θ)= sec

^{4}θ – sec^{2}θ= R.H.S

Hence proved .

**Question 7: What will be the supplement of sin, tan, and cos.**

**Solution:**

- Supplement angle of sin(180° – x) = sin x
- Supplement angle of tan(180° – x) = -tan x
- Supplement angle of cos180° – x) = -cos x