Question

$\huge{\underline{\underline{\mathfrak{\sf{Question-}}}}}$


If $\sf{\dfrac{Sin^4\:\theta}{a}}+{\dfrac{Cos^4\:\theta}{b}}={\dfrac{1}{a+b}}$, then prove that $\sf{\dfrac{Sin^8\:\theta}{a^3}}+{\dfrac{Cos^8\:\theta}{b^3}}={\dfrac{1}{(a+b)^3}}$


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Answer 1

$\bf{\Huge{\boxed{\sf{\blue{ANSWER\::}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

$\tt{\frac{sin^{4}\theta }{a} \:+\frac{cos^{4} \theta}{b} \:=\:\frac{1}{a+b} }$

$\bf{\Large{\underline{\bf{To\:prove\::}}}}$

$\rm{\frac{sin^{8}\theta }{a^{3} } \:+\frac{cos^{8} \theta}{b^{3} }\:=\:\frac{1}{(a+b)^{3} } }$

$\bf{\Large{\boxed{\sf{\red{Explanation\::}}}}}}$

We know that,

$\leadsto\sf{sin^{2} \theta\:+\:cos^{2} \theta\:=\:1}$

$\leadsto\sf{cos^{2} \theta\;=\:1-sin^{2} \theta......................(1)}$

Therefore,

$\longmapsto\tt{\frac{sin^{4} \theta}{a} +\frac{(cos^{2}a)^{2} \theta }{b} =\frac{1}{a+b} }$

Putting the value of equation (1) in place of cos²Ф, we get;

$\longmapsto\tt{\frac{sin^{4}\theta }{a} +\frac{(1-sin^{2}\theta)^{2}  }{b} =\frac{1}{a+b} }$

$\bf{\large{\boxed{\rm{\pink{Using\:(a\:-\:b)^{2} =a^{2} +b^{2}-2ab \::}}}}}}$

$\longmapsto\tt{\frac{sin^{4}\theta }{a}\:+\frac{(1)^{2}+(sin^{2} \theta) ^{2} -2*1*sin^{2} \theta}{b} =\frac{1}{a+b} }$

$\longmapsto\tt{\frac{sin^{4} \theta}{a} \:+\frac{1+sin^{4}\theta-2sin^{2} \theta }{b} =\frac{1}{a+b} }$

$\longmapsto\tt{bsin^{4} \theta\:+a+asin^{4} \theta\:-2asin^{2} \theta\:=\:\frac{ab}{a+b} }$

$\longmapsto\tt{sin^{4} \theta(a+b)-2asin^{2} +a=\frac{ab}{a+b} }$

$\longmapsto\tt{sin^{4} \theta(a+b)^{2} -2asin^{2} \theta(a+b)+a(a+b)=ab}$

$\longmapsto\tt{sin^{4} \theta(a+b)^{2} -2asin^{2} \theta(a+b)+a^{2}+\cancel{ab}\cancel{-ab}=0}$

$\longmapsto\tt{[sin^{2} \theta(a+b)]^{2} -2asin^{2} \theta(a+b)+a^{2} =0}$

$\longmapsto\tt{[sin^{2} \theta(a+b)-a]=0}$

$\longmapsto\tt{sin^{2} \theta(a+b)=a}$

$\longmapsto\tt{\orange{sin^{2} \theta\:=\:\frac{a}{a+b}} }$

&

Putting the value of sin²Ф in equation (1), we get;

$\longmapsto\tt{cos^{2}\theta\:=\:1-\frac{a}{a+b} }$

$\longmapsto\tt{cos^{2}\theta\:=\:\frac{\cancel{a}+b\cancel{-a}}{a+b}  }$

$\longmapsto\tt{\orange{cos^{2} \theta\:=\:\frac{b}{a+b} }}$

Now,

$\longmapsto\tt{\frac{sin^{8}\theta }{a^{3} } \:+\frac{cos^{8}\theta }{b^{3} } =\frac{1}{(a+b)^{3} } }$

$\longmapsto\tt{\frac{(sin^{2}\theta )^{4} }{a^{3} }\: +\:\frac{(cos^{2}\theta)^{4} }{b^{3} } =\frac{1}{(a+b)^{3} } }$

$\longmapsto\tt{\frac{\cancel{a^{4}} }{(a+b)^{4} *\cancel{a^{3}} }\:+\frac{\cancel{b^{4}} }{(a+b)^{4} *\cancel{b^{3} }} =\frac{1}{(a+b)^{3} } }$

$\longmapsto\tt{\frac{a}{(a+b)^{4} } \:+\:\frac{b}{(a+b)^{4} } \:=\:\frac{1}{(a+b)^{3} } }$

$\longmapsto\tt{\frac{a+b}{(a+b)^{4} } \:=\:\frac{1}{(a+b)^{3} } }$

$\longmapsto\tt{\frac{1}{(a+b)^{3} } \:=\:\frac{1}{(a+b)^{3} }}$

Hence,

Proved. ♥

Answer 2

Answer:

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