 Fiona Petersen

2022-01-26

If ${\mathrm{cos}}^{4}x.{\mathrm{sec}}^{2}y,\frac{1}{2},{\mathrm{sin}}^{4}x.{\mathrm{csc}}^{2}y$ are in A.P, then prove that ${\mathrm{cos}}^{8}x.{\mathrm{sec}}^{6}y,\frac{1}{2},{\mathrm{sin}}^{8}x.{\mathrm{csc}}^{6}y$ in A.P
My Attempt
${\mathrm{cos}}^{4}x.{\mathrm{sec}}^{2}y+{\mathrm{sin}}^{4}.x{\mathrm{csc}}^{2}y=\frac{{\mathrm{cos}}^{4}x}{{\mathrm{cos}}^{2}y}+\frac{{\mathrm{sin}}^{4}x}{{\mathrm{sin}}^{2}y}=1$
$⇒{\mathrm{sin}}^{2}y.{\mathrm{cos}}^{4}x+{\mathrm{cos}}^{2}y.{\mathrm{sin}}^{4}x={\mathrm{sin}}^{2}y.{\mathrm{cos}}^{2}y$
${\mathrm{cos}}^{8}x.{\mathrm{sec}}^{6}y+{\mathrm{sin}}^{8}x.{\mathrm{csc}}^{6}y=\frac{{\mathrm{cos}}^{8}x}{{\mathrm{cos}}^{6}y}+\frac{{\mathrm{sin}}^{8}x}{{\mathrm{sin}}^{6}y}$
How do I know that the given terms are in A.P, G.P or H.P ? Karli Kaiser

Expert

Hint:
$\frac{12}{-}{\mathrm{cos}}^{4}x=2\left({\mathrm{sin}}^{4}x-\frac{1}{2}\right)$
$⇔2-{\left(1+\mathrm{cos}2x\right)}^{2}=2{\left(1-\mathrm{cos}2x\right)}^{2}-4$
Solve for $\mathrm{cos}2x$
${\mathrm{csc}}^{2}y-\frac{1}{2}=2\left(\frac{1}{2}-{\mathrm{sec}}^{2}y\right)$
$⇔2-4\left(1+{\mathrm{tan}}^{2}y\right)=2\left(1+{\mathrm{cot}}^{2}y\right)-1$
$4{\mathrm{tan}}^{2}y+2\cdot \frac{1}{{\mathrm{tan}}^{2}y}-5=0$
Solve for ${\mathrm{tan}}^{2}y\ge 0$

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