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Trả lời:
$d,$ Ta có:
$VP=\sin^2x.\cot^2x=\sin^2x.\dfrac{\cos^2x}{\sin^2x}=\cos^2x$
$=1-\sin^2x=(1-\sin x)(1+\sin x)=VT$
Vậy $(1-\sin x)(1+\sin x)=\sin^2x.\cot^2x$.
$1,$ Ta có:
$VP=\cos^2a(1-\tan a)+\sin^2a(1-\cot a)$
$=\cos^2a-\cos^2a.\dfrac{\sin a}{\cos a}+\sin^2a-\sin^2a.\dfrac{\cos a}{\sin a}$
$=\sin^2a+\cos^2-2\sin a\cos a$
$=(\sin a-\cos)^2=VT$
Vậy $(\sin a-\cos a)^2=\cos^2a(1-\tan a)+\sin^2a(1-\cot a)$.
