§2. Giá trị lượng giác của một cung

\(VT=tana+\frac{cosa}{1+sina}=\frac{sina}{cosa}+\frac{cosa}{1+sina}\)

\(=\frac{sina+sin^2a+cos^2a}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)

\(P=\frac{sina}{cosa}\left(\frac{1+cos^2a}{sina}-sina\right)=\frac{1}{cosa}\left(1+cos^2a-sin^2a\right)\)

\(=\frac{1}{cosa}\left(cos^2a+cos^2a\right)=\frac{2cos^2a}{cosa}=2cosa\)

\(=\dfrac{sina}{cosa}+\dfrac{1+cos^2a-sin^2a}{sina}\)

\(=\dfrac{sin^2a+2cos^3a}{sina\cdot cosa}\)

a/ \(\frac{sinx+cotx}{1+sinx.tanx}=\frac{sinx.cosx\left(sinx+cotx\right)}{sinx.cosx\left(1+sinx.tanx\right)}=\frac{cosx\left(sin^2x+cosx\right)}{sinx\left(cosx+sin^2x\right)}=cotx\)

\(\Rightarrow VT=cot^{2013}x\)

\(VP=\frac{sin^{2013}x+cot^{2013}x}{1+sin^{2013}x.tan^{2013}x}=\frac{sin^{2013}x.cos^{2013}x\left(sin^{2013}x+cot^{2013}x\right)}{sin^{2013}x.cos^{2013}x\left(1+sin^{2013}x.tan^{2013}x\right)}\)

\(=\frac{cos^{2013}x\left(sin^{4026}x+cos^{2013}x\right)}{sin^{2013}x\left(cos^{2013}x+sin^{4036}x\right)}=\frac{cos^{2013}x}{sin^{2013}x}=cot^{2013}x=VT\) (đpcm)

b/ \(\left(\sqrt{\frac{1+sinx}{1-sinx}}-\sqrt{\frac{1-sinx}{1+sinx}}\right)^2=\frac{1+sinx}{1-sinx}+\frac{1-sinx}{1+sinx}-2\)

\(=\frac{\left(1+sinx\right)^2+\left(1-sinx\right)^2}{\left(1-sinx\right)\left(1+sinx\right)}-2=\frac{2+2sin^2x}{1-sin^2x}-2=\frac{2+2sin^2x}{cos^2x}-2\)

\(=\frac{2}{cos^2x}-2+2tan^2x=\frac{2\left(1-cos^2x\right)}{cos^2x}+2tan^2x=2tan^2x+2tan^2x=4tan^2x\)

\(\dfrac{x+y}{t+z}=\dfrac{3}{5}\)

=>5x+5y=3t+3z

=>5x-3t=3z-5y=0

=>5x=3t

=>x/t=3/5

Lời giải:

Ta có:

VT\(=\frac{1+\cot ^2x}{1-\cot ^2x}+\frac{\cos x}{\cos x-\sin x}=\frac{1+\left(\frac{\cos x}{\sin x}\right)^2}{1-\left(\frac{\cos x}{\sin x}\right)^2}+\frac{\cos x}{\cos x-\sin x}\)

\(=\frac{\sin ^2x+\cos ^2x}{\sin ^2x(1-\frac{\cos ^2x}{\sin ^2x})}+\frac{\cos x(\cos x+\sin x)}{\cos ^2x-\sin ^2x}\)

\(=\frac{1}{\sin ^2x-\cos ^2x}-\frac{\cos x(\cos x+\sin x)}{\sin ^2x-\cos ^2x}\)

\(=\frac{1-\cos ^2x-\cos x\sin x}{\sin ^2x-\cos ^2x}=\frac{\sin ^2x-\cos x\sin x}{\sin ^2x-\cos ^2x}\)

\(=\frac{\sin x(\sin x-\cos x)}{\sin ^2x-\cos ^2x}=\frac{\sin x}{\sin x+\cos x}\)

Ta có đpcm.

\(1+\cot^2a=\dfrac{1}{\sin^2a}=1+\dfrac{1}{4}=\dfrac{5}{4}\)

\(\Leftrightarrow\sin^2a=\dfrac{4}{5}\)

hay \(\sin a=-\dfrac{2\sqrt{5}}{5}\left(\Pi< a< \dfrac{3\Pi}{2}\right)\)

=>\(\cos a=-\dfrac{\sqrt{5}}{5}\)

\(\sin^2a\cdot\cos a=\dfrac{4}{5}\cdot\dfrac{-\sqrt{5}}{5}=\dfrac{-4\sqrt{5}}{25}\)

Chọn B

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