Đáp án:
Giải thích các bước giải:
g) \(VT=\dfrac{\sqrt{2}cos\ a-2cos(\dfrac{\pi}{4}+a)}{2sin(\dfrac{\pi}{4}+a)-\sqrt{2}sin\ a}\)
\(VT=\dfrac{\sqrt{2}cos\ a-2[cos\ \dfrac{\pi}{4}. cos\ a-sin\ \dfrac{\pi}{4} . sin\ a]}{2(sin\ \dfrac{\pi}{4}.cos\ a+cos\ \dfrac{\pi}{4}.sin\ a)-\sqrt{2}sin\ a}\)
\(VT=\dfrac{\sqrt{2}cos\ a-\sqrt{2}cos\ a+\sqrt{2}sin\ a}{\sqrt{2}cos\ a+\sqrt{2}sin\ a-\sqrt{2}sin\ a}\)
\(VT=\dfrac{\sqrt{2}sin\ a}{\sqrt{2}cos\ a }=tan\ a=VP\)
`⇒ ĐPCM`
h) `VT=\frac{cos 2a}{1+sin 2a}`
`VT=\frac{sin^2 a+cos^2 a-2sin a. cos a}{sin^2 a+cos^2 a+2sin a . cos a}`
`VT=\frac{(cos a-sin a)^2}{(cos a+sin a)^2}`
\(VT=\dfrac{[\sqrt{2}(cos^2.\dfrac{1}{\sqrt{2}}-sin\ a.\dfrac{1}{\sqrt{2}})]^2}{[\sqrt{2}(cos\ a . sin\ \dfrac{\pi}{4}+sin\ a.cos\ \dfrac{\pi}{4})]^2}\)
\(VT=\dfrac{\sqrt{2}cos^2(a+\dfrac{\pi}{4})}{\sqrt{2}sin^2(a+\dfrac{\pi}{4})}\)
\(VT=cot(\dfrac{\pi}{4}+a)=VP\)
`⇒ ĐPCM`
