Brooks Mckenzie
02/21/2023 · Junior High School
\( \frac { \tan ^ { 2 } A } { 1 \operatorname { tax } ^ { 2 } A } : \sin ^ { 2 } A \)
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Step-by-step Solution
To simplify the expression \( \frac{\tan^2 A}{1 \operatorname{tax}^2 A} : \sin^2 A \), we first need to clarify the notation. It seems there might be a typo with "operatorname{tax}". Assuming you meant "tan", we can rewrite the expression as:
\[
\frac{\tan^2 A}{1 + \tan^2 A} : \sin^2 A
\]
The colon ":" indicates division, so we can rewrite it as:
\[
\frac{\tan^2 A}{1 + \tan^2 A} \div \sin^2 A
\]
This can be expressed as:
\[
\frac{\tan^2 A}{1 + \tan^2 A} \cdot \frac{1}{\sin^2 A}
\]
Now, recall the identity for tangent:
\[
\tan A = \frac{\sin A}{\cos A} \implies \tan^2 A = \frac{\sin^2 A}{\cos^2 A}
\]
Substituting this into our expression gives:
\[
\frac{\frac{\sin^2 A}{\cos^2 A}}{1 + \frac{\sin^2 A}{\cos^2 A}} \cdot \frac{1}{\sin^2 A}
\]
Now, simplify the denominator:
\[
1 + \frac{\sin^2 A}{\cos^2 A} = \frac{\cos^2 A + \sin^2 A}{\cos^2 A} = \frac{1}{\cos^2 A}
\]
Thus, we can rewrite the expression as:
\[
\frac{\frac{\sin^2 A}{\cos^2 A}}{\frac{1}{\cos^2 A}} \cdot \frac{1}{\sin^2 A}
\]
This simplifies to:
\[
\sin^2 A \cdot \frac{1}{\sin^2 A} = 1
\]
Therefore, the final simplified result is:
\[
\boxed{1}
\]
Quick Answer
The expression simplifies to 1.
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