\(a,\sqrt{2}sin\left(\alpha+\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha cos\dfrac{\pi}{4}+cos\alpha sin\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha\cdot\dfrac{\sqrt{2}}{2}+cos\alpha\cdot\dfrac{\sqrt{2}}{2}\right)-cos\alpha\\ =\sqrt{2}\cdot sin\alpha\cdot\dfrac{\sqrt{2}}{2}+\sqrt{2}\cdot cos\alpha\cdot\dfrac{\sqrt{2}}{2}-cos\alpha\\ =sin\alpha+cos\alpha-cos\alpha\\ =sin\alpha\)

\(b,\left(cos\alpha+sin\alpha\right)^2-sin2\alpha\\ =cos^2\alpha+sin^2\alpha=2cos\alpha sin\alpha-2sin\alpha cos\alpha\\ =sin^2\alpha+cos^2\alpha\\ =1\)

\(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=4\\ VT=\dfrac{sin^2a+2sinacosa+cos^2a-sin^2a+2sinacosa-cos^2a}{sinacosa}\\ =\dfrac{4sinacosa}{sinacosa}=4=VP\)

a: \(S=cos^2a\left(1+tan^2a\right)=cos^2a\cdot\dfrac{1}{cos^2a}=1\)

b: \(VP=\dfrac{1+sin2a-1+sin2a}{\dfrac{1}{2}\cdot sin2a}=\dfrac{2\cdot sin2a}{\dfrac{1}{2}\cdot sin2a}=4=VT\)

a) S= \(cos^2a\left(tg^2a+1\right)=cos^2a.\dfrac{1}{cos^2a}=1\)

\(A=\dfrac{cos^2a-sin^2a}{\dfrac{cos^2a}{sin^2a}-\dfrac{sin^2a}{cos^2a}}-cos^2a=\dfrac{cos^2a.sin^2a\left(cos^2a-sin^2a\right)}{\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)}-cos^2a\)

\(=cos^2a.sin^2a-cos^2a=cos^2a\left(sin^2a-1\right)=-cos^4a\)

\(B=\sqrt{\left(1-cos^2a\right)^2+6cos^2a+3cos^4a}+\sqrt{\left(1-sin^2a\right)^2+6sin^2a+3sin^4a}\)

\(=\sqrt{4cos^4a+4cos^2a+1}+\sqrt{4sin^4a+4sin^2a+1}\)

\(=\sqrt{\left(2cos^2a+1\right)^2}+\sqrt{\left(2sin^2a+1\right)^2}\)

\(=2\left(sin^2a+cos^2a\right)+2=4\)

\(a,\dfrac{1}{tan\alpha+1}+\dfrac{1}{cot\alpha+1}\\ =\dfrac{cot\alpha+1+tan\alpha+1}{\left(tan\alpha+1\right)\left(cot\alpha+1\right)}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha\cdot cot\alpha+tan\alpha+cot\alpha+1}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha+cot\alpha+2}\\ =1\)

\(b,cos\left(\dfrac{\pi}{2}-\alpha\right)-sin\left(\pi+\alpha\right)\\ =sin\alpha+sin\alpha\\ =2sin\alpha\)

\(c,sin\left(\alpha-\dfrac{\pi}{2}\right)+cos\left(-\alpha+6\pi\right)-tan\left(\alpha+\pi\right)cot\left(3\pi-\alpha\right)\\ =-sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\alpha\right)-tan\left(\alpha\right)cot\left(\pi-\alpha\right)\\ =-cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\alpha\right)\cdot cot\left(\alpha\right)\\ =1\)

Mẫu số là \(-3cos2a\) hay \(-2cos2a\) vậy bạn? -3 không hợp lý

a) \(\dfrac{\sin2\text{a}+\cos a}{1+\cos2\text{a}+\cos a}=2\tan a\)

a) \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos\alpha}=\dfrac{2sin\alpha cos\alpha+sin\alpha}{2cos^2\alpha+cos\alpha}\)\(=\dfrac{sin\alpha\left(2cos\alpha+1\right)}{cos\alpha\left(2cos\alpha+1\right)}=\dfrac{sin\alpha}{cos\alpha}=tan\alpha\).

b) \(\dfrac{4sin^2\alpha}{1-cos^2\dfrac{\alpha}{2}}=\dfrac{4sin^2\alpha}{sin^2\dfrac{\alpha}{2}}=\dfrac{4.sin^2\dfrac{\alpha}{2}.cos^2\dfrac{\alpha}{2}}{sin^2\dfrac{\alpha}{2}}=4sin^2\dfrac{\alpha}{2}\).

\(\frac{1}{cos^2a}=1+tan^2a\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{1}{10}\)

a/ \(\frac{sina-cosa}{sina+cosa}=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}=\frac{3-1}{3+1}\)

b/ \(\frac{2sina+3cosa}{3sina-5cosa}=\frac{3tana+3}{3tana-5}=\frac{3.3+3}{3.3-5}\)

c/ \(\frac{1+2cos^2a}{1-cos^2a-cos^2a}=\frac{1+2cos^2a}{1-2cos^2a}=\frac{1+2.\frac{1}{10}}{1-2.\frac{1}{10}}\)

d/ \(\frac{\left(1-cos^2a\right)^2+\left(cos^2a\right)^2}{1+1-cos^2a}=\frac{\left(1-\frac{1}{10}\right)^2+\left(\frac{1}{10}\right)^2}{2-\frac{1}{10}}\)

\(\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cos2a+sin3a}{2cos3a.cos2a+cos3a}=\dfrac{sin3a\left(2cos2a+1\right)}{cos3a\left(2cos2a+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)

\(\dfrac{1+sin4a-cos4a}{1+sin4a+cos4a}=\dfrac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin2a.cos2a+2cos^22a-1}=\dfrac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\dfrac{sin2a}{cos2a}=tan2a\)

\(96\sqrt{3}sin\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=48\sqrt{3}sin\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)\)

\(=24\sqrt{3}sin\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=12\sqrt{3}sin\left(\dfrac{\pi}{6}\right)cos\left(\dfrac{\pi}{6}\right)\)

\(=6\sqrt{3}sin\left(\dfrac{\pi}{3}\right)=6\sqrt{3}.\dfrac{\sqrt{3}}{2}=9\)

\(A+B+C=\pi\Rightarrow A+B=\pi-C\Rightarrow tan\left(A+B\right)=tan\left(\pi-C\right)\)

\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=-tanC\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)

\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\)