sin4+cos4=1-2sin^2cos^2
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Answer:
Step-by-step explanation:
\frac{sin^{4}\theta+cos^{4}\theta}{1-2sin^{2}\theta cos^{2}\theta}=1
Step-by-step explanation:
LHS= \frac{sin^{4}\theta+cos^{4}\theta}{1-2sin^{2}\theta cos^{2}\theta}\\=\frac{(sin^{2}\theta)^{2}+(cos^{2}\theta)^{2}}{1-2sin^{2}\theta cos^{2}\theta}\\=\frac{(sin^{2}\theta+cos^{2}\theta)^{2}-2sin^{2}\theta cos^{2}\theta }{1-2sin^{2}\theta cos^{2}\theta}\\
/* By algebraic identity:
i )a²+b² = (a+b)²-2ab
By Trigonometric identity:
ii) sin²A+cos²A = 1 */
= \frac{1-2sin^{2}\theta cos^{2}\theta }{1-sin^{2}\theta cos^{2}\theta }\\=1 \\=RHS
Therefore,
\frac{sin^{4}\theta+cos^{4}\theta}{1-2sin^{2}\theta cos^{2}\theta}=1
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