Question

[Thầy Tùng Dương]

Chứng minh các đẳng thức sau(giả sử các biểu thức sau đều có nghĩa)
a) $\sin ^{4} x+\cos ^{4} x=1-2 \sin ^{2} x \cdot \cos ^{2} x$.

b) $\dfrac{1+\cot x}{1-\cot x}=\dfrac{\tan x+1}{\tan x-1}$.

c) $\dfrac{\cos x+\sin x}{\cos ^{3} x}=\tan ^{3} x+\tan ^{2} x+\tan x+1$.

Answer 1

[Nguyễn Xuân Anh]

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

\(\Leftrightarrow sin^4x+2sin^2x\cdot cos^2x+cos^4x=1\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2=1\)(luôn đúng)

Answer 2

[Cao Thị Kim Ngân]

a) ${\sin }^{4}x+{\cos }^{4}x={\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x-2{\sin }^{2}x{\cos }^{2}x$
\begin{aligned}&=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x \\&=1-2 \sin ^{2} x \cos ^{2} x\end{aligned}

b) $\genfrac{}{}{0ex}{}{1+\cot x}{1-\cot x}=\genfrac{}{}{0ex}{}{1+\genfrac{}{}{0ex}{}{1}{\tan x}}{1-\genfrac{}{}{0ex}{}{1}{\tan x}}=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{\tan x+1}{\tan x}}{\genfrac{}{}{0ex}{}{\tan x-1}{\tan x}}=\genfrac{}{}{0ex}{}{\tan x+1}{\tan x-1}$

c) $\genfrac{}{}{0ex}{}{\cos x+\sin x}{{\cos }^{3}x}=\genfrac{}{}{0ex}{}{1}{{\cos }^{2}x}+\genfrac{}{}{0ex}{}{\sin x}{{\cos }^{3}x}={\tan }^{2}x+1+\tan x\left({\tan }^{2}x+1\right)$
$={\tan }^{3}x+{\tan }^{2}x+\tan x+1$