If \(\cos A+\cos^2A=1\), then \(\sin^2A+\sin^4A=1\).
True.
Step 1: Let \(x = \cos A\). The given condition is:
\(x + x^2 = 1\).
Step 2: Rearrange to find \(x^2\):
\(x^2 = 1 - x\).
Step 3: Recall that \(\sin^2 A = 1 - \cos^2 A\).
So, \(\sin^2 A = 1 - x^2\).
Step 4: Now calculate \(\sin^2 A + \sin^4 A\):
\(\sin^2 A + \sin^4 A = (1 - x^2) + (1 - x^2)^2\).
Step 5: Expand the square:
\((1 - x^2)^2 = 1 - 2x^2 + x^4\).
So, \(\sin^2 A + \sin^4 A = (1 - x^2) + (1 - 2x^2 + x^4)\).
Step 6: Simplify:
\(\sin^2 A + \sin^4 A = 2 - 3x^2 + x^4\).
Step 7: Replace \(x^2\) using Step 2: \(x^2 = 1 - x\).
So, \(x^4 = (x^2)^2 = (1 - x)^2 = 1 - 2x + x^2\).
But \(x^2 = 1 - x\), so:
\(x^4 = 1 - 2x + (1 - x) = 2 - 3x\).
Step 8: Substitute this back:
\(2 - 3x^2 + x^4 = 2 - 3(1 - x) + (2 - 3x)\).
Step 9: Simplify step by step:
= \(2 - 3 + 3x + 2 - 3x\)
= \(1\).
Final Result: Therefore, \(\sin^2 A + \sin^4 A = 1\). The statement is True.