Question

Plz solve 9 and 10 both . Who will do it correctly I will mark it the brainliest​

Answer 1

QUESTION 1:

$Putting \: \:z = \pi/4 \: \: show \: \: that \\ \\ \displaystyle \lim_{x \to\pi/4} \dfrac{1 - \tan x}{1 - \sqrt{2} \sin x} = 2$

FORMULAE USED:

$\tan(x + y) = \dfrac{ \tan x + \tan y}{1 - \tan x \tan y} \\ \\ \\\sin (x + y) = \sin x \cos y+\cos x \sin y \\ \\ \\ \tan \bigg( \dfrac{ \pi}{4} \bigg) = 1 \\ \\ \\ \sin\bigg( \dfrac{ \pi}{4} \bigg) = \dfrac{1}{ \sqrt{2} } \\ \\ \\ \cos \bigg( \dfrac{ \pi}{4} \bigg) = \dfrac{1}{ \sqrt{2} } \\ \\ \\ 1 - \cos x = 2 { \sin}^{2} x \\ \\ \\ sinx = 2 \sin( \frac{x}{2} ) \cos( \frac{x}{2} ) \\ \\ \\ \dfrac{ \sin x }{x} = 1 \\ \\ \\ \dfrac{ \tan x }{x} = 1$

Tan 0° = 0

Sin 0° = 0

cos 0° = 1

QUESTION 2:

$Putting \: \: x = \theta - \frac{ \pi}{4} \: prove \: \: that \\ \\ \displaystyle \lim_{x \to\pi/4} \dfrac{ \sin \theta - \cos \theta}{ \theta - \dfrac{ \pi}{4}}$

FORMULAE USED:

$\sin (x + y) = \sin x \cos y+\cos x \sin y \\ \\ \\ \cos(x + y) = \cos x\cos y - \sin x\sin y \\ \\ \\ \sin\bigg( \dfrac{ \pi}{4} \bigg) = \dfrac{1}{ \sqrt{2} } \\ \\ \\ \cos \bigg( \dfrac{ \pi}{4} \bigg) = \dfrac{1}{ \sqrt{2} } \\ \\ \\ \dfrac{ \sin x}{x} = 1$

NOTE:REFER ATTACHMENT FOR PROOF.