Chứng minh đẳng thức sau: $sin^2x(1+cotx)+cos^2x(1+tanx)=(sinx+cosx)^2$ 29/08/2021 Bởi Parker Chứng minh đẳng thức sau: $sin^2x(1+cotx)+cos^2x(1+tanx)=(sinx+cosx)^2$
$VT=\sin^2x(1+\cot x)+\cos^2x(1+\tan x)$ $=\sin^2x+\cos^2x+\sin^2x\cot x+\cos^2x\tan x$ $=\sin^2x+\cos^2x+\sin x\cos x+\sin x\cos x$ $=\sin^2x+2\sin x\cos x+\cos^2x$ $=(\sin x+\cos x)^2$ $=VP$ Bình luận
$\begin{array}{l}{\sin ^2}x\left( {1 + \cot x} \right) + {\cos ^2}x\left( {1 + \tan x} \right)\\ = {\sin ^2}x + {\sin ^2}x.\cot x + {\cos ^2}x + {\cos ^2}.\tan x\\ = {\sin ^2}x + {\cos ^2}x + {\sin ^2}x\frac{{\cos x}}{{\sin x}} + {\cos ^2}x\frac{{\sin x}}{{\cos x}}\\ = {\sin ^2}x + {\cos ^2}x + 2\sin x.\cos x = {\left( {\sin x + \cos x} \right)^2}\end{array}$ Bình luận
$VT=\sin^2x(1+\cot x)+\cos^2x(1+\tan x)$
$=\sin^2x+\cos^2x+\sin^2x\cot x+\cos^2x\tan x$
$=\sin^2x+\cos^2x+\sin x\cos x+\sin x\cos x$
$=\sin^2x+2\sin x\cos x+\cos^2x$
$=(\sin x+\cos x)^2$
$=VP$
$\begin{array}{l}
{\sin ^2}x\left( {1 + \cot x} \right) + {\cos ^2}x\left( {1 + \tan x} \right)\\
= {\sin ^2}x + {\sin ^2}x.\cot x + {\cos ^2}x + {\cos ^2}.\tan x\\
= {\sin ^2}x + {\cos ^2}x + {\sin ^2}x\frac{{\cos x}}{{\sin x}} + {\cos ^2}x\frac{{\sin x}}{{\cos x}}\\
= {\sin ^2}x + {\cos ^2}x + 2\sin x.\cos x = {\left( {\sin x + \cos x} \right)^2}
\end{array}$