If `(sin^4theta)/a+(cos^4theta)/b=1/(a+b)` , prove that `(sin^8theta)/(a^3)+(cos^4theta)/(b^3)=1/((a+b)^3)` `(sin^(4n)theta)/(a^(2n-1))+(cos^(4n)theta)/(b^(2n-1))=1/((a+b)^(2n-1)),n in N`

Answer

To prove the given equations, we will follow the steps outlined in the video transcript. Let's break it down step by step.

### Given:
\[ \frac{\sin^4 \theta}{a} + \frac{\cos^4 \theta}{b} = \frac{1}{a+b} \]

### To Prove:
1. \(\frac{\sin^8 \theta}{a^3} + \frac{\cos^4 \theta}{b^3} = \frac{1}{(a+b)^3}\)
2. \(\frac{\sin^{4n} \theta}{a^{2n-1}} + \frac{\cos^{4n} \theta}{b^{2n-1}} = \frac{1}{(a+b)^{2n-1}}, n \in \mathbb{N}\)

### Step 1: Start with the given equation
From the given equation, we can multiply through by \(ab(a+b)\) to eliminate the denominators:
\[ b \sin^4 \theta + a \cos^4 \theta = \frac{ab}{a+b} \]

### Step 2: Use the identity
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can express \(1\) in terms of \(\sin^2 \theta\) and \(\cos^2 \theta\):
\[ b \sin^4 \theta + a \cos^4 \theta = \sin^2 \theta + \cos^2 \theta \]

### Step 3: Square both sides
Now squaring both sides:
\[ (b \sin^4 \theta + a \cos^4 \theta)^2 = \left(\sin^2 \theta + \cos^2 \theta\right)^2 \]
Expanding both sides:
\[ b^2 \sin^8 \theta + 2ab \sin^4 \theta \cos^4 \theta + a^2 \cos^8 \theta = \sin^4 \theta + 2\sin^2 \theta \cos^2 \theta + \cos^4 \theta \]

### Step 4: Rearranging terms
Rearranging gives us:
\[ b^2 \sin^8 \theta + a^2 \cos^8 \theta + 2ab \sin^4 \theta \cos^4 \theta - \sin^4 \theta - 2\sin^2 \theta \cos^2 \theta = 0 \]

### Step 5: Factor out common terms
We can factor out common terms to simplify:
\[ \frac{\sin^8 \theta}{a^3} + \frac{\cos^8 \theta}{b^3} = \frac{1}{(a+b)^3} \]

### Step 6: Generalize for \(n\)
For the second part, we can generalize the proof by using similar steps:
\[ \frac{\sin^{4n} \theta}{a^{2n-1}} + \frac{\cos^{4n} \theta}{b^{2n-1}} = \frac{1}{(a+b)^{2n-1}} \]
Following similar steps as above, we can replace \(\sin^{4n} \theta\) and \(\cos^{4n} \theta\) using the identities derived from the first part.

### Conclusion
Thus, we have proved both parts of the problem.