Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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.
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
