Câu 1:

a: Đúng

b: \(A=\sin^2x+3\cdot\sin x\cdot cosx-4\cdot cos^2x\)

\(=1-cos^2x-4\cdot cos^2x+3\cdot\sin x\cdot cosx\)

\(=1-5\cdot cos^2x+3\cdot\sin x\cdot cosx\)

=>\(\frac{1-5\cdot cos^2x+3\cdot\sin x\cdot cosx}{cos^2x}=\frac{1}{cos^2x}-5+3\cdot\frac{\sin x}{cosx}\)

\(=\tan^2x+1-5+3\cdot\tan x=\tan^2x+3\cdot\tan x-4\)

=>\(A\cdot\left(\tan^2x+1\right)=\tan^2x+3\cdot\tan x-4\)

=>\(A=\frac{\tan^2x+3\cdot\tan x-4}{\tan^2x+1}\)

=>Đúng

c: \(P=\frac{\sin^2x+3\cdot\sin x\cdot cosx-4\cdot cos^2x}{\tan x-1}\)

\(=\frac{\tan^2x+3\cdot\tan x-4}{\tan^2x+1}:\left(\tan x-1\right)=\frac{\left(\tan x+4\right)\left(\tan x-1\right)}{\left(\tan x-1\right)\left(\tan^2x+1\right)}=\frac{\tan x+4}{\tan^2x+1}\)

=>Đúng

d: \(\frac{1}{cos^2x}=\tan^2x+1\)

=>\(\tan^2x+1=\frac{1}{\left(\frac12\right)^2}=1:\frac14=4\)

=>\(\tan^2x=3\)

=>\(tanx=\sqrt3\) hoặc \(tanx=-\sqrt3\)

\(P=\frac{\tan x+4}{1+\tan^2x}=\frac{\tan x+4}{4}\)

Khi tan x=\(\sqrt3\) thì \(P=\frac{4+\sqrt3}{4}\)

Khi tan x=-\(\sqrt3\) thì \(P=\frac{4-\sqrt3}{4}\)

=>Sai

Câu 2:

a: \(\left(\sin x+cosx\right)^2=\sin^2x+cos^2x+2\cdot\sin x\cdot cosx\)

\(=1+2\cdot\sin x\cdot cosx\)

=>Đúng

b: \(\tan^2x-\sin^2x\)

\(=\frac{\sin^2x}{cos^2x}-\sin^2x=\sin^2x\left(\frac{1}{cos^2x}-1\right)\)

\(=\sin^2x\cdot\frac{1-cos^2x}{cos^2x}=\sin^2x\cdot\frac{\sin^2x}{cos^2x}=\sin^2x\cdot\tan^2x\)

=>Đúng

c: Sai

d: \(A=\frac{\tan^2x-\sin^2x+\left(\sin x+cosx\right)^2-1}{\tan^2x\cdot\sin^2x}\)

\(=\frac{\tan^2x\cdot\sin^2x-2\cdot\sin x\cdot cosx}{\tan^2x\cdot\sin^2x}=1-\frac{2}{\sin x}\cdot\frac{cosx}{\tan^2x}=1-\frac{2}{\sin x}\cdot\frac{cosx\cdot cos^2x}{\sin^2x}\)

\(=1-\frac{2\cdot cos^3x}{\sin^3x}=1-2\cdot\cot^3x\)

=>Sai