Question

$\sf{Chapter:}$ Inverse Trigonometric Functions

$\sf{Class:}$ 12th

$\sf{Note:}$ By Substitution Method​

Answer 1

Hey there❤

Refer to attachment

Answer 2

Step-by-step explanation:

(i)

$Given:\lim_{x \to a} \frac{\sqrt{tanx} - \sqrt{tana}}{x - a}$

On rationalizing the numerator, we get

$=\frac{\sqrt{tanx} - \sqrt{tana}}{x - a}*\frac{\sqrt{tanx}+\sqrt{tana}}{\sqrt{tanx} +\sqrt{tana}}$

$=\frac{tanx-tana}{(x-a)(\sqrt{tanx} + \sqrt{tana})}$

$\boxed{tanA - tanB = tan(A - B)[1 +tanAtanB]}$

$=\frac{tan(x-a)(1+tanxtana)}{(x-a)(\sqrt{tanx} + \sqrt{tana})}$

$=\lim_{x \to a} \frac{1+tanatana}{\sqrt{tana} + \sqrt{tana}}$

$=\frac{1+tan^2a}{2\sqrt{tana}}$

$=\frac{sec^2a}{2\sqrt{tana} }$

(ii)

$Given:\lim_{x \to 2} \frac{x^2 - x - 2}{x^3 - 7}$

$=\frac{2^2-2-2}{2^3-7}$

$=\frac{4-2-2}{8-7}$

$=\frac{0}{1}$

$=0$

Hope it helps!