# Prove that (1 – sin A)/(1 + sin A) = (sec A – tan A)²

Trigonometry name itself says that it is a subject that deals with the geometry of triangles and it is very useful for situations when needed to find when there are some sides given and we need the relations between the sides or angles between the sides. In Trigonometry we have different ratios that are sin A, cos A, tan A, cot A, sec A, cosec A with the help of which, the relation between the sides and the angle between the sides of the triangle can be obtained.

### Trigonometric functions

The trigonometric functions define the relation between the sides and angles and the examples are sin A, cos A, tan A, cot A, sec A, cosec A. The relation between different trigonometric functions is a** trigonometric identity**. The identities are very useful to test the inequality in the trigonometric equations. Examples are,

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- Tan A = sin A/cos A
- sin A = 1/cosec A
- cos A = 1/sec A
- Tan A = 1/cot A

### Prove that (1 – sin A)/(1 + sin A) = (sec A – tan A)²

There are basic identities that are required in order to solve the above problem statement, lets look at some of the basic identities of the 6 trigonometric functions that are required in this case,

**Prerequisite Identities used in the proof**

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

1 – sin^{2}A = cos^{2}A - tan A = sin A/cos A
- sec A = 1/cos A
- sin A = 1/cosec A
- (a – b)
^{2 }= a^{2 }– 2 × a × b + b^{2} - a
^{2 }– b^{2 }= (a + b)(a – b) - (a – b)/c = a/c – b/c
- sin 2A = 2 × sin A × cos A

**Given Trigonometric equation**

(1 – sin A)/(1 + sin A) = (sec A – tan A)^{2}

LHS = (1 – sin A)/(1 + sin A)

RHS = (sec A – tan A)^{2}

**Deriving Proof from LHS side**

Given LHS(1 – sin A)/(1 + sin A)

Step-1Multiplying with (1 – sin A)/(1 – sin A) which is equal to 1 to bring the degree of 2 which is present on RHS.

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

=(1 – sin A)

^{2}/(1 – sin^{2}A)

Step-2Expanding the numerator (1 – sinA)

^{2}=(1 – 2 × sin A + sin

^{2}A)/cos^{2}A

Step-3Breaking the equation in step-2 in general form

=(1/cos

^{2}A) – 2(sin A/cos A)(1/cos A) + (sin^{2}A/cos^{2}A)

Step-4Substituting with standard formulas in the equation obtained in step-3

=(sec

^{2}A) – 2(tan A) × (sec A) + tan^{2}A=(sec A – Tan A)

^{2}

From step-4 it can be concluded that LHS = (sec A – Tan A)^{2} which is equal to RHS and thus,

(sec A – Tan A)^{2} = (sec A- Tan A)^{2 }

LHS = RHS

**Hence Proved.**

**Deriving Proof from the RHS side**

Given RHS(sec A – Tan A)

^{2}

Step-1Simplifying the equation by substituting standard formulas=((1/cos A) – ( sin A/cos A))

^{2}=((1 – sin A)/cos A)

^{2}

Step 2Simplifying the denominator=((1 – sin A)(1 – sin A))/(1 – sin

^{2}A)=((1 – sin A)(1 – sin A))/( (1 – sin A)(1 + sin A))

Step-3(1 – sin A) in numerator and denominator of the equation in step-2 gets cancelled so it becomes

=(1 – sin A)/(1 + sin A)

From the step-3 it can be concluded that RHS = (1 – sin A)/(1+ sin A) which is equal to LHS and thus,

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

LHS = RHS

**Hence Proved.**

### Sample Problems

**Question 1: Solve the trigonometric identity: ((cosec A – 1)/(cosec A+1)) × ((1 – sin ^{2}A)/(1 – sinA)^{2})**

**Solution:**

- By using identity 4 in the equation
= {((1/sin A) – 1)/((1/sin A) + 1)) × ((1 – sin

^{2}A)/(1 – sinA)^{2}}

- By using identity 1 in the equation
= ((1 – sin A)/(1 + sin A)) × ((cosA)

^{2}/(1 – sin A)^{2})

- Multiplying and dividing by cos
^{2}A= ((1 – sin A)/(1 + sin A)) × (1/(sec A – Tan A)

^{2})

- By using our derived identity
= (sec A – Tan A)

^{2}x (1/(sec A – Tan A)^{2})= 1

**Question 2: Solve the trigonometric identity: ((sec A/2) + 2sinA/2)/((sec A/2) – 2sinA/2) ) × (4/(sec A – Tan A) ^{2})**

**Solution:**

- By using identity-3
= 4 × ((1/(cos A/2) + 2sinA/2))/((1/(cos A/2) – 2sinA/2)) × (1/(sec A – Tan A)

^{2})

- By using the identity-8
= 4 × ((1+ 2 sin A/2 × cos A/2)/(1 – 2 sin A/2 × cos A/2)) × (1/(sec A – Tan A)

^{2})= 4 × ((1 + sin A )/(1 – sin A)) × (1/(sec A – Tan A)

^{2})

- By using the identity proved
= 4 × (sec A – Tan A)

^{2 }× (1/(sec A – Tan A)^{2})= 4