Let $f: R \rightarrow R: f(x)=x^{2}, g: R \rightarrow R: g(x)=\tan x$
and $h: R \rightarrow R: h(x)=\log x .$
Find a formula for h o (g o f).
Show that [h o (g of)] $\sqrt{\frac{\pi}{4}}=0$
To find: formula for h o (g o f)
To prove:
Formula used: $f$ o $f=f(f(x))$
Given: (i) $f: R \rightarrow R: f(x)=x^{2}$
(ii) $g: R \rightarrow R: g(x)=\tan x$
(iii) $h: R \rightarrow R: h(x)=\log x$
Solution: We have,
ho $(g \circ f)=h \circ g(f(x))=h \circ g\left(x^{2}\right)$
$=h\left(g\left(x^{2}\right)\right)=h\left(\tan x^{2}\right)$
$=\log \left(\tan x^{2}\right)$
h o $(g \circ f)=\log \left(\tan x^{2}\right)$
For, $[h \circ(g \circ f)] \sqrt{\frac{\pi}{4}}$
$=\log \left[\tan \left(\sqrt{\frac{\pi}{4}}\right)^{2}\right]$
$=\log \left[\tan \frac{\pi}{4}\right]$
$=\log 1$
= 0
Hence Proved.
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