Question

. The function f(x) = tan x - x
(a) always increases
(b) always decreases
(c) never increases
(d) sometimes increases and sometimes decreases
explanation please​

Answer 1

☯ Explanation,

Fuction, f(x) = tan x - x

On differentiating w.r.t x,

⇒ f'(x) = d/dx (tan (x) - x)

Using differentiation rule,

{ °.° d/dx(f + g) = d/dx(f) + d/dx(g) }

⇒ f'(x) = d/dx(tan(x) - d/dx(x)

⇒ f'(x) = sec x² - x

{ °.° sec²x - x = tan²x }

⇒ f'(x) = tan²x

{ °.° Square of any number is always greater than zero and always tends to increase. }

⇒ f'(x) = tan²x ≥ 0

☯ Hence,

The given fuction is always increasing.

Answer 2

Option A

Always increasing

$f(x)=tanx−x$

$f′(x)=sec(x)2−1>0$

Or ,

$sec(x)2>1$

Now 

$cos(x)ϵ[−1,1]$

Hence

$sec(x)ϵ(−∞,−1]∪[1,∞).$

Thus 

$sec2(x)ϵ[1,∞)$

Hence

$f′(x)>0 for all x.$

Hence

f(x) is always increasing.