Câu hỏi của 2003

Question

chứng minh rằng

2) $sin^2x+sin^2\left(\frac{\pi}{2}-x\right)-tan\left(\frac{\pi}{2}+x\right)tanx=2$

Nguyễn Việt Lâm · Accepted Answer

$sin^2x+sin^2\left(\frac{\pi}{2}-x\right)-tan\left(\frac{\pi}{2}+x\right).tanx$

$=sin^2x+cos^2x-\left(-cotx\right).tanx$

$=1-\left(-1\right)=2$Câu trả lời: $sin^2x+sin^2\left(\frac{\pi}{2}-x\right)-tan\left(\frac{\pi}{2}+x\right).tanx$

$=sin^2x+cos^2x-\left(-cotx\right).tanx$

$=1-\left(-1\right)=2$