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,

• 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

• sin²A + cos²A = 1
  1 – sin²A = cos²A
• tan A = sin A/cos A
• sec A = 1/cos A
• sin A = 1/cosec A
• (a – b)² = a² – 2 × a × b + b²
• a² – b² = (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-1

Multiplying 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 – sin²A)

Step-2

Expanding the numerator (1 – sinA)²

=(1 – 2 × sin A + sin²A)/cos²A

Step-3

Breaking the equation in step-2 in general form 

=(1/cos²A) – 2(sin A/cos A)(1/cos A) + (sin²A/cos²A)

Step-4

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

=(sec ²A) – 2(tan A) × (sec A) + tan² 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-1 Simplifying the equation by substituting standard formulas

=((1/cos A) – ( sin A/cos A))²

=((1 – sin A)/cos A)²

Step 2 Simplifying the denominator

=((1 – sin A)(1 – sin A))/(1 – sin²A)

=((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 – sin²A)/(1 – sinA)²)

Solution:

• By using identity 4 in the equation

= {((1/sin A) – 1)/((1/sin A) + 1)) × ((1 – sin²A)/(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 cos²A

= ((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