Làm nhanh giúp e nhé !!!

Câu 19:

1) 

$\sin x=\sqrt{1-\cos^2x}=0,6$

$\tan x=\dfrac{\sin x}{\cos x}=0,75$

$\cot x=\dfrac{1}{\tan x}=\dfrac{4}{3}$

2)

$\cot x=\dfrac{1}{\tan x}=\dfrac{4}{3}$

$\dfrac{1}{\cos^2x}=1+\tan^2x\Rightarrow \cos x=0,8$

$\sin x=\sqrt{1-\cos^2x}=0,6$

3)

$\sin x=\cos x$

Với $0<x<90^o$, $\sin x>0, \cos x>0$

$\Leftrightarrow \sin^2x=\cos^2x$

$\Leftrightarrow \sin^2x=1-\sin^2x$

$\Leftrightarrow \sin x=\dfrac{1}{\sqrt2}$

$\cos x=\dfrac{1}{\sqrt2}$

$\tan x=\dfrac{\sin x}{\cos x}=1$

$\cot x=\dfrac{1}{\tan x}=1$

4)

$\tan x=\cot x=\dfrac{1}{\tan x}$

$\Leftrightarrow \tan^2x=1$

$\Leftrightarrow \tan x=\cot x=1$

$\dfrac{1}{\cos^2x}=1+\tan^2x$

$\Rightarrow \cos x=\dfrac{1}{\sqrt2}$

$\sin x=\sqrt{1-\cos^2x}=\dfrac{1}{\sqrt2}$

Câu 20:

Chia cả tử và mẫu của C cho $\cos x$:

$C=\dfrac{1+\tan x}{1-\tan x}=\dfrac{1+2}{1-2}=-3$

$\cos x=\dfrac{1}{3}\Leftrightarrow cos^2x=\dfrac{1}{9}$

$P=3(1-\cos^2x)+\cos^2x$

$=3-2\cos^2x=\dfrac{25}{9}$