The expression (1 + tan x + {tan ^2}x) (1 - c | vedclass

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Question

(c) The expression is $\frac{{(1 + 	an x + {{	an }^2}x)(1 + {{	an }^2}x - 	an x)}}{{{{	an }^2}x}}$

$=  \frac{{{{(1 + {{	an }^2}x)}^2} -

Vedclass · Accepted Answer

For all $x \in R$

Vedclass · Answer

(c) The expression is $\frac{{(1 + 	an x + {{	an }^2}x)(1 + {{	an }^2}x - 	an x)}}{{{{	an }^2}x}}$

$=  \frac{{{{(1 + {{	an }^2}x)}^2} - {{	an }^2}x}}{{{{	an }^2}x}}$

Obviously, $1 + {	an ^2}x \ge {	an ^2}x,{m{ }}\forall {m{ }}x$. 

Hence it is positive for all value of $x.$

The expression $(1 + \tan x + {\tan ^2}x)$ $(1 - \cot x + {\cot ^2}x)$ has the positive values for $x$, given by

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