Answer :
Sure, let's analyze and address the given trigonometric equation step-by-step:
We are given the equation:
[tex]\[ \sqrt{\frac{1 - \cos A}{1 + \cos A}} + \sqrt{\frac{1 + \cos A}{1 - \cos A}} = 2 \csc A \][/tex]
To solve this, let's define each term and see if we can simplify or analyze the equation. Here's the breakdown:
1. Left-hand Side Simplification (LHS):
[tex]\[ \text{LHS} = \sqrt{\frac{1 - \cos A}{1 + \cos A}} + \sqrt{\frac{1 + \cos A}{1 - \cos A}} \][/tex]
First, let's introduce:
[tex]\[ x = \sqrt{\frac{1 - \cos A}{1 + \cos A}} \][/tex]
Then, the second term can be written as:
[tex]\[ \sqrt{\frac{1 + \cos A}{1 - \cos A}} = \frac{1}{x} \][/tex]
So, the left-hand side becomes:
[tex]\[ x + \frac{1}{x} \][/tex]
2. Right-hand Side Simplification (RHS):
[tex]\[ \text{RHS} = 2 \csc A = \frac{2}{\sin A} \][/tex]
3. Equating LHS and RHS:
We need to check if:
[tex]\[ x + \frac{1}{x} = \frac{2}{\sin A} \][/tex]
4. Analysis:
For the equality:
[tex]\[ x + \frac{1}{x} = \frac{2}{\sin A} \][/tex]
Define:
- [tex]\(\sin A = y\)[/tex]
- [tex]\(\cos A = \sqrt{1 - y^2}\)[/tex]
If we substitute [tex]\(\cos A \)[/tex] in [tex]\( x\)[/tex], we get:
[tex]\[ x = \sqrt{\frac{1 - \sqrt{1 - y^2}}{1 + \sqrt{1 - y^2}}} \][/tex]
Algebraic manipulation can become quite complex. Let's analyze using different values of [tex]\( A \)[/tex] (angles).
For specific angles:
1. When [tex]\( A = 0^\circ\)[/tex]
- [tex]\(\cos 0^\circ = 1\)[/tex]
- [tex]\(\sin 0^\circ = 0\)[/tex]
This could lead to undefined expressions on RHS (since division by zero is undefined).
2. When [tex]\( A = 90^\circ \)[/tex]
- [tex]\(\cos 90^\circ = 0\)[/tex]
- [tex]\(\sin 90^\circ = 1\)[/tex]
Let's check the LHS:
[tex]\[ \sqrt{\frac{1 - 0}{1 + 0}} + \sqrt{\frac{1 + 0}{1 - 0}} = 1 + 1 = 2 \][/tex]
RHS:
[tex]\[ 2 \csc 90^\circ = 2 \times 1 = 2 \][/tex]
Thus, for [tex]\( 90^\circ \)[/tex], both LHS and RHS agree.
However, for most values of [tex]\( A \)[/tex], the given evaluations like at [tex]\( 0^\circ \)[/tex] and closer angles will become inconsistent, suggesting it won't hold for all values of [tex]\( A \)[/tex].
From this, we conclude:
[tex]\[ \sqrt{\frac{1 - \cos A}{1 + \cos A}} + \sqrt{\frac{1 + \cos A}{1 - \cos A}} = 2 \csc A \][/tex]
is not true for all values of [tex]\( A \)[/tex].
Hence, the equation:
[tex]\[ \sqrt{\frac{1-\cos A}{1+\cos A}}+\sqrt{\frac{1+\cos A}{1-\cos A}}=2 \operatorname{cosec} A \][/tex]
is false in general.
We are given the equation:
[tex]\[ \sqrt{\frac{1 - \cos A}{1 + \cos A}} + \sqrt{\frac{1 + \cos A}{1 - \cos A}} = 2 \csc A \][/tex]
To solve this, let's define each term and see if we can simplify or analyze the equation. Here's the breakdown:
1. Left-hand Side Simplification (LHS):
[tex]\[ \text{LHS} = \sqrt{\frac{1 - \cos A}{1 + \cos A}} + \sqrt{\frac{1 + \cos A}{1 - \cos A}} \][/tex]
First, let's introduce:
[tex]\[ x = \sqrt{\frac{1 - \cos A}{1 + \cos A}} \][/tex]
Then, the second term can be written as:
[tex]\[ \sqrt{\frac{1 + \cos A}{1 - \cos A}} = \frac{1}{x} \][/tex]
So, the left-hand side becomes:
[tex]\[ x + \frac{1}{x} \][/tex]
2. Right-hand Side Simplification (RHS):
[tex]\[ \text{RHS} = 2 \csc A = \frac{2}{\sin A} \][/tex]
3. Equating LHS and RHS:
We need to check if:
[tex]\[ x + \frac{1}{x} = \frac{2}{\sin A} \][/tex]
4. Analysis:
For the equality:
[tex]\[ x + \frac{1}{x} = \frac{2}{\sin A} \][/tex]
Define:
- [tex]\(\sin A = y\)[/tex]
- [tex]\(\cos A = \sqrt{1 - y^2}\)[/tex]
If we substitute [tex]\(\cos A \)[/tex] in [tex]\( x\)[/tex], we get:
[tex]\[ x = \sqrt{\frac{1 - \sqrt{1 - y^2}}{1 + \sqrt{1 - y^2}}} \][/tex]
Algebraic manipulation can become quite complex. Let's analyze using different values of [tex]\( A \)[/tex] (angles).
For specific angles:
1. When [tex]\( A = 0^\circ\)[/tex]
- [tex]\(\cos 0^\circ = 1\)[/tex]
- [tex]\(\sin 0^\circ = 0\)[/tex]
This could lead to undefined expressions on RHS (since division by zero is undefined).
2. When [tex]\( A = 90^\circ \)[/tex]
- [tex]\(\cos 90^\circ = 0\)[/tex]
- [tex]\(\sin 90^\circ = 1\)[/tex]
Let's check the LHS:
[tex]\[ \sqrt{\frac{1 - 0}{1 + 0}} + \sqrt{\frac{1 + 0}{1 - 0}} = 1 + 1 = 2 \][/tex]
RHS:
[tex]\[ 2 \csc 90^\circ = 2 \times 1 = 2 \][/tex]
Thus, for [tex]\( 90^\circ \)[/tex], both LHS and RHS agree.
However, for most values of [tex]\( A \)[/tex], the given evaluations like at [tex]\( 0^\circ \)[/tex] and closer angles will become inconsistent, suggesting it won't hold for all values of [tex]\( A \)[/tex].
From this, we conclude:
[tex]\[ \sqrt{\frac{1 - \cos A}{1 + \cos A}} + \sqrt{\frac{1 + \cos A}{1 - \cos A}} = 2 \csc A \][/tex]
is not true for all values of [tex]\( A \)[/tex].
Hence, the equation:
[tex]\[ \sqrt{\frac{1-\cos A}{1+\cos A}}+\sqrt{\frac{1+\cos A}{1-\cos A}}=2 \operatorname{cosec} A \][/tex]
is false in general.
