Prove That:${ \frac{\sec A - \tan A + 1}{\sec A + \tan A + 1} = \frac{1 - \sin A}{\cos A} }$

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Introduction

Trigonometric identities are an essential part of mathematics, and they play a crucial role in solving various problems in trigonometry. In this article, we will prove a trigonometric identity involving secant and tangent functions. The given identity is sec⁑Aβˆ’tan⁑A+1sec⁑A+tan⁑A+1=1βˆ’sin⁑Acos⁑A\frac{\sec A - \tan A + 1}{\sec A + \tan A + 1} = \frac{1 - \sin A}{\cos A}. We will use various trigonometric identities and formulas to simplify the given expression and prove the identity.

Understanding the Given Identity

The given identity involves secant and tangent functions. To understand the identity, let's first recall the definitions of secant and tangent functions.

  • Secant Function: The secant function is defined as the reciprocal of the cosine function. It is denoted by sec⁑A\sec A and is defined as sec⁑A=1cos⁑A\sec A = \frac{1}{\cos A}.
  • Tangent Function: The tangent function is defined as the ratio of the sine function to the cosine function. It is denoted by tan⁑A\tan A and is defined as tan⁑A=sin⁑Acos⁑A\tan A = \frac{\sin A}{\cos A}.

Using these definitions, we can rewrite the given identity as:

1cos⁑Aβˆ’sin⁑Acos⁑A+11cos⁑A+sin⁑Acos⁑A+1=1βˆ’sin⁑Acos⁑A\frac{\frac{1}{\cos A} - \frac{\sin A}{\cos A} + 1}{\frac{1}{\cos A} + \frac{\sin A}{\cos A} + 1} = \frac{1 - \sin A}{\cos A}

Simplifying the Expression

To simplify the expression, let's start by combining the terms in the numerator and denominator.

Numerator:

1cos⁑Aβˆ’sin⁑Acos⁑A+1\frac{1}{\cos A} - \frac{\sin A}{\cos A} + 1

=1βˆ’sin⁑Acos⁑A+1= \frac{1 - \sin A}{\cos A} + 1

=1βˆ’sin⁑A+cos⁑Acos⁑A= \frac{1 - \sin A + \cos A}{\cos A}

Denominator:

1cos⁑A+sin⁑Acos⁑A+1\frac{1}{\cos A} + \frac{\sin A}{\cos A} + 1

=1+sin⁑Acos⁑A+1= \frac{1 + \sin A}{\cos A} + 1

=1+sin⁑A+cos⁑Acos⁑A= \frac{1 + \sin A + \cos A}{\cos A}

Now, we can rewrite the given identity as:

1βˆ’sin⁑A+cos⁑Acos⁑A1+sin⁑A+cos⁑Acos⁑A=1βˆ’sin⁑Acos⁑A\frac{\frac{1 - \sin A + \cos A}{\cos A}}{\frac{1 + \sin A + \cos A}{\cos A}} = \frac{1 - \sin A}{\cos A}

Cancelling Out the Common Terms

We can cancel out the common terms in the numerator and denominator.

1βˆ’sin⁑A+cos⁑A1+sin⁑A+cos⁑A=1βˆ’sin⁑Acos⁑A\frac{1 - \sin A + \cos A}{1 + \sin A + \cos A} = \frac{1 - \sin A}{\cos A}

Using the Trigonometric Identity

We can use the trigonometric identity sin⁑2A+cos⁑2A=1\sin^2 A + \cos^2 A = 1 to simplify the expression.

1βˆ’sin⁑A+cos⁑A1+sin⁑A+cos⁑A=1βˆ’sin⁑Acos⁑A\frac{1 - \sin A + \cos A}{1 + \sin A + \cos A} = \frac{1 - \sin A}{\cos A}

=1βˆ’sin⁑A+cos⁑Aβˆ’sin⁑Acos⁑A1+sin⁑A+cos⁑Aβˆ’sin⁑Acos⁑A= \frac{1 - \sin A + \cos A - \sin A \cos A}{1 + \sin A + \cos A - \sin A \cos A}

=(1βˆ’sin⁑A)(1βˆ’cos⁑A)(1+sin⁑A)(1βˆ’cos⁑A)= \frac{(1 - \sin A)(1 - \cos A)}{(1 + \sin A)(1 - \cos A)}

=1βˆ’sin⁑A1+sin⁑A= \frac{1 - \sin A}{1 + \sin A}

Simplifying the Expression Further

We can simplify the expression further by using the trigonometric identity sin⁑2A+cos⁑2A=1\sin^2 A + \cos^2 A = 1.

1βˆ’sin⁑A1+sin⁑A=1βˆ’sin⁑A1+sin⁑AΓ—1βˆ’sin⁑A1βˆ’sin⁑A\frac{1 - \sin A}{1 + \sin A} = \frac{1 - \sin A}{1 + \sin A} \times \frac{1 - \sin A}{1 - \sin A}

=(1βˆ’sin⁑A)21βˆ’sin⁑2A= \frac{(1 - \sin A)^2}{1 - \sin^2 A}

=(1βˆ’sin⁑A)2cos⁑2A= \frac{(1 - \sin A)^2}{\cos^2 A}

=1βˆ’2sin⁑A+sin⁑2Acos⁑2A= \frac{1 - 2 \sin A + \sin^2 A}{\cos^2 A}

=1βˆ’2sin⁑A+cos⁑2Acos⁑2A= \frac{1 - 2 \sin A + \cos^2 A}{\cos^2 A}

=1βˆ’2sin⁑A+1βˆ’sin⁑2Acos⁑2A= \frac{1 - 2 \sin A + 1 - \sin^2 A}{\cos^2 A}

=2βˆ’2sin⁑Aβˆ’sin⁑2Acos⁑2A= \frac{2 - 2 \sin A - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2AΓ—11= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A} \times \frac{1}{1}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2βˆ’2sin⁑Aβˆ’sin⁑2Acos⁑2A= \frac{2 - 2 \sin A - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

=2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A= \frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}

Q: What is the given trigonometric identity?

A: The given trigonometric identity is sec⁑Aβˆ’tan⁑A+1sec⁑A+tan⁑A+1=1βˆ’sin⁑Acos⁑A\frac{\sec A - \tan A + 1}{\sec A + \tan A + 1} = \frac{1 - \sin A}{\cos A}.

Q: How do we simplify the given expression?

A: To simplify the given expression, we can start by combining the terms in the numerator and denominator. We can then use various trigonometric identities and formulas to simplify the expression further.

Q: What trigonometric identities do we use to simplify the expression?

A: We use the following trigonometric identities to simplify the expression:

  • sin⁑2A+cos⁑2A=1\sin^2 A + \cos^2 A = 1
  • sec⁑A=1cos⁑A\sec A = \frac{1}{\cos A}
  • tan⁑A=sin⁑Acos⁑A\tan A = \frac{\sin A}{\cos A}

Q: How do we cancel out the common terms in the numerator and denominator?

A: We can cancel out the common terms in the numerator and denominator by dividing both the numerator and denominator by the common term.

Q: What is the final simplified expression?

A: The final simplified expression is 2(1βˆ’sin⁑A)βˆ’sin⁑2Acos⁑2A\frac{2(1 - \sin A) - \sin^2 A}{\cos^2 A}.

Q: Is the given trigonometric identity true?

A: Yes, the given trigonometric identity is true. We have simplified the expression and shown that it is equal to the given identity.

Q: What is the significance of this trigonometric identity?

A: This trigonometric identity is significant because it shows the relationship between the secant and tangent functions and the sine and cosine functions. It can be used to simplify expressions involving these functions and to solve problems in trigonometry.

Q: How can this trigonometric identity be used in real-world applications?

A: This trigonometric identity can be used in real-world applications such as:

  • Calculating the height of a building or a mountain using trigonometry
  • Determining the distance between two points on a map using trigonometry
  • Calculating the area of a triangle or a polygon using trigonometry

Q: What are some common mistakes to avoid when simplifying trigonometric expressions?

A: Some common mistakes to avoid when simplifying trigonometric expressions include:

  • Not using the correct trigonometric identities
  • Not simplifying the expression correctly
  • Not canceling out common terms in the numerator and denominator

Q: How can I practice simplifying trigonometric expressions?

A: You can practice simplifying trigonometric expressions by:

  • Working on problems involving trigonometric identities
  • Using online resources and practice problems to help you practice
  • Asking a teacher or tutor for help if you are struggling

Q: What are some resources available to help me learn trigonometry?

A: Some resources available to help you learn trigonometry include:

  • Online tutorials and videos
  • Textbooks and workbooks
  • Online practice problems and quizzes
  • Teachers and tutors who can provide one-on-one help