Question

plz answer the 9th and 10th questions with steps friends

Answer 1

9)
as you know
$cos(2x) = 2cos^{2}(x)-1$
so
[tex]cos(2x) + cos^{2}(x) = 3cos^{2}(x)-1 \\ [/tex]
and
$-1 \leq cos(x) \leq 1$
$0 \leq cos^{2} (x) \leq 1$
$-1 \leq 3cos^{2} (x) - 1 \leq 2$
Hence extreme values are -1, 2

10)
as you know
$cosh (x) = \frac{e^{x}+e^{-x}}{2}$
So
[tex]2 cosh^{2} (x) - 1 = 2(\frac{e^{x}+e^{-x}}{2})^{2} - 1 \\ = \frac{e^{2x}+e^{-2x}+2}{2} - 1 =\frac{e^{2x}+e^{-2x}}{2} = cosh (2x)[/tex]