Question

if u give correct answer with solution,I mark you as brainliest


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Answer 1

Answer:

Mean value theorem states that if a function f(x) is continuous and differentiable in interval (a,b), then,

f(b)−f(a)b−a=f'(c), where c lies in (a,b).

Let f(x)=tanx where x∈(0,π2) .

Then, from Lagranges Mean value theorem,

f'(c)=f(b)−f(a)b−a

If we apply Mean value theorem on [0,x].

Then, sec2c=tanx−tan0x−0

⇒sec2c=tanxx→(1)

Here, 0<c<x

⇒sec20<sec2c<sec2x

From (1),

⇒1<tanxx<sec2x

⇒1<tanxx

⇒x<tanx

⇒tanx>x.

Step-by-step explanation: follow and give thnx plz