Answer

Verified

340.5k+ views

**Hint:**Here, we will use the Trigonometric Identity, trigonometric ratios and a suitable Algebraic Identity to prove the given trigonometric function. Trigonometric Ratios of a Particular angle are the ratios of the sides of a right angled triangle with respect to any of its acute angle.

**Formula Used:**

We will use the following formula:

1. Trigonometric Identity: \[{\sin ^2}x + {\cos ^2}x = 1\].

2. The difference between the square of the numbers is given by an Algebraic Identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\].

3. Trigonometric Ratio: \[\csc A = \dfrac{1}{{\sin A}}\] and \[\cot A = \dfrac{{\cos A}}{{\sin A}}\].

**Complete Step by Step Solution:**

We are given that the trigonometric function \[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \csc A + \cot A\].

Now, we will multiply the numerator and the denominator of the given fraction by \[\sin A\], so we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\sin A\left( {\cos A - \sin A + 1} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

Now, by multiplying the terms with the terms inside the Parentheses only in the numerator, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\sin A\cos A - {{\sin }^2}A + \sin A}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

Trigonometric Identity: \[{\sin ^2}x + {\cos ^2}x = 1\]

Now, by rewriting the equation using the Trigonometric Identity, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\sin A\cos A + \sin A - \left( {1 - {{\cos }^2}A} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

The difference between the square of the numbers is given by an Algebraic Identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Now, by using an Algebraic Identity, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\sin A\cos A + \sin A - \left( {1 + \cos A} \right)\left( {1 - \cos A} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

Now, by taking out the common terms, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\sin A\left( {1 + \cos A} \right) - \left( {1 + \cos A} \right)\left( {1 - \cos A} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

Again by taking out the common terms, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\left( {1 + \cos A} \right)\left( {\sin A - 1 + \cos A} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\left( {1 + \cos A} \right)\left( {\cos A + \sin A - 1} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

Now, by cancelling out the common terms, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\left( {1 + \cos A} \right)}}{{\sin A}}\]

Now, by segregating the terms, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{1}{{\sin A}} + \dfrac{{\cos A}}{{\sin A}}\]

Trigonometric Ratio: \[\csc A = \dfrac{1}{{\sin A}}\] and \[\cot A = \dfrac{{\cos A}}{{\sin A}}\]

Now, by using the Trigonometric Ratio, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \csc A + \cot A\]

**Therefore, \[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \csc A + \cot A\]is proved.**

**Note:**

We know that Trigonometric Equation is defined as an equation involving trigonometric ratios. Trigonometric identity is an equation which is always true for all the variables. We should know that we have many trigonometric identities which are related to all the other trigonometric equations. Trigonometric Ratios are used to find the relationships between the sides of a right angle triangle.

Recently Updated Pages

The branch of science which deals with nature and natural class 10 physics CBSE

Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Distinguish between the reserved forests and protected class 10 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give simple chemical tests to distinguish between the class 12 chemistry CBSE

Difference Between Plant Cell and Animal Cell

Which of the following books is not written by Harshavardhana class 6 social science CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

In which states of India are mango showers common What class 9 social science CBSE

What Made Mr Keesing Allow Anne to Talk in Class class 10 english CBSE