If 10 ^4 +15 ^4 =6, then the value of 27 ^6 +8 ^6 16 ^8 is

Question

If $10{\sin }^{4}	heta +15{\cos }^{4}	heta =6$, then the value of $\frac{27{\csc }^{6}	heta +8{\sec }^{6}	heta }{16{\sec }^{8}	heta }$ is

Accepted Answer

$\begin{gathered} 10{\sin }^{4}	heta +15{\cos }^{4}	heta =6 \ \Rightarrow 10{\sin }^{4}	heta +10{\cos }^{4}	heta +5{\cos }^{4}	heta =6 \ \Rightarrow 10[{({\sin }^{2}	heta +{\cos }^{2}	heta )}^{2}-2{\sin }^{2}	heta {\cos }^{2}	heta ]+5{\cos }^{4}	heta =6 \ \Rightarrow 10-20(1-{\cos }^{2}	heta ){\cos }^{2}	heta +5{\cos }^{4}	heta =6 \ {\cos }^{2}	heta =x \ 10-20(x-{x}^{2})+5{x}^{2}=6 $

$ \Rightarrow 25{x}^{2}-20x+4=0 \ (5x-2{)}^{2}=0\Rightarrow x=\frac{2}{5} \ \Rightarrow {\cos }^{2}	heta =\frac{2}{5}\Rightarrow {\sin }^{2}	heta =\frac{3}{5}, \ {\sec }^{2}	heta =\frac{5}{2},{\csc }^{2}	heta =\frac{5}{3} $

$ \frac{27{\csc }^{6}	heta +8{\sec }^{6}	heta }{16{\sec }^{8}	heta } =\frac{27{(\frac{5}{3})}^{3}+8{(\frac{5}{2})}^{3}}{16{(\frac{5}{2})}^{4}} \ =\frac{{5}^{3}+{5}^{3}}{{5}^{4}}=\frac{{2.5}^{3}}{{5}^{4}}=\frac{2}{5} \end{gathered}$

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