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Trigonometric Relation (Tomato Subjective 115)

Problem: If \(\displaystyle { \frac{\sin^4 x }{a} + \frac{\cos^4 x }{b} = \frac{1}{a+b} }\) , then show that \(\displaystyle { \frac{\sin^6 x }{a^2} + \frac{\cos^6 x }{b^2} = \frac{1}{(a+b)^2} }\)

Discussion: 

Put \(\sin^2 x = t \).

The given expression reduces to \(\displaystyle { \frac{t^2 }{a} + \frac{1+t^2 -2t }{b} = \frac{1}{a+b} }\)

\(\displaystyle {\Rightarrow (a+b)t^2 -2at +a – \frac{ab}{a+b} = 0 }\)
\(\displaystyle {\Rightarrow (a+b)^2t^2 -2a(a+b)t +a^2 = 0 }\)
\(\displaystyle {\Rightarrow ((a+b)t-a)^2=0 }\)
\(\displaystyle {\Rightarrow \frac{a}{a+b}=t }\)
\(\displaystyle {\Rightarrow \frac{b}{a+b}=1-t }\)

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November 26, 2015

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