Garza Bradley
08/05/2024 · High School
Prove that: \( \frac{\tan ^{3} \alpha}{1+\tan ^{2} \alpha}+\frac{\cot ^{2} \alpha}{1+\cot ^{2} \alpha}=\sec \alpha \operatorname{cosec} \alpha-2 \sin \alpha \cos \alpha \)
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Step-by-step Solution
To prove the identity
\[
\frac{\tan^3 \alpha}{1 + \tan^2 \alpha} + \frac{\cot^2 \alpha}{1 + \cot^2 \alpha} = \sec \alpha \csc \alpha - 2 \sin \alpha \cos \alpha,
\]
we start by simplifying the left-hand side.
### Step 1: Simplify the left-hand side
We know that
\[
\tan \alpha = \frac{\sin \alpha}{\cos \alpha} \quad \text{and} \quad \cot \alpha = \frac{\cos \alpha}{\sin \alpha}.
\]
Thus, we can express \(\tan^2 \alpha\) and \(\cot^2 \alpha\) as follows:
\[
\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha}, \quad \cot^2 \alpha = \frac{\cos^2 \alpha}{\sin^2 \alpha}.
\]
Now, we can rewrite the first term:
\[
\frac{\tan^3 \alpha}{1 + \tan^2 \alpha} = \frac{\frac{\sin^3 \alpha}{\cos^3 \alpha}}{1 + \frac{\sin^2 \alpha}{\cos^2 \alpha}} = \frac{\frac{\sin^3 \alpha}{\cos^3 \alpha}}{\frac{\cos^2 \alpha + \sin^2 \alpha}{\cos^2 \alpha}} = \frac{\sin^3 \alpha \cos^2 \alpha}{\cos^3 \alpha (\cos^2 \alpha + \sin^2 \alpha)}.
\]
Since \(\cos^2 \alpha + \sin^2 \alpha = 1\), this simplifies to:
\[
\frac{\tan^3 \alpha}{1 + \tan^2 \alpha} = \frac{\sin^3 \alpha \cos^2 \alpha}{\cos^3 \alpha}.
\]
Next, we simplify the second term:
\[
\frac{\cot^2 \alpha}{1 + \cot^2 \alpha} = \frac{\frac{\cos^2 \alpha}{\sin^2 \alpha}}{1 + \frac{\cos^2 \alpha}{\sin^2 \alpha}} = \frac{\frac{\cos^2 \alpha}{\sin^2 \alpha}}{\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin^2 \alpha}} = \frac{\cos^2 \alpha}{\sin^2 \alpha} \cdot \frac{\sin^2 \alpha}{1} = \cos^2 \alpha.
\]
### Step 2: Combine the terms
Now we combine both simplified terms:
\[
\frac{\tan^3 \alpha}{1 + \tan^2 \alpha} + \frac{\cot^2 \alpha}{1 + \cot^2 \alpha} = \frac{\sin^3 \alpha \cos^2 \alpha}{\cos^3 \alpha} + \cos^2 \alpha.
\]
To combine these, we can express \(\cos^2 \alpha\) with a common denominator:
\[
\cos^2 \alpha = \frac{\cos^5 \alpha}{\cos^3 \alpha}.
\]
Thus, we have:
\[
\frac{\sin^3 \alpha \cos^2 \alpha + \cos^5 \alpha}{\cos^3 \alpha}.
\]
### Step 3: Simplify the numerator
The numerator can be factored:
\[
\sin^3 \alpha \cos^2 \alpha + \cos^5 \alpha = \cos^2 \alpha (\sin^3 \alpha + \cos^3 \alpha).
\]
Using the identity \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\), we have:
\[
\sin^3 \alpha + \cos^3 \alpha = (\sin \alpha + \cos \alpha)(\sin^2 \alpha - \sin \alpha \cos \alpha + \cos^2 \alpha) = (\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha).
\]
Thus, the left-hand side becomes:
\[
\frac{\cos^2 \alpha (\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha)}{\cos^3 \alpha} = \frac{(\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha)}{\cos \alpha}.
Quick Answer
The left-hand side simplifies to \(\frac{(\sin \alpha + \cos \alpha)(1 - \sin \alpha \cos \alpha)}{\cos \alpha}\).
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