a.
\(=\dfrac{2sina.cosa+sina}{1+\left(2cos^2a-1\right)+cosa}=\dfrac{sina\left(2cosa+1\right)}{cosa\left(2cosa+1\right)}=\dfrac{sina}{cosa}=tana\)
b.
\(=\dfrac{\left(2sina.cosa\right)^2-4\left(1-sin^2a\right)}{1-8sin^2a-\left(1-2sin^22a\right)}=\dfrac{4sin^2a.cos^2a-4cos^2a}{2\left(2.sina.cosa\right)^2-8sin^2a}\)
\(=\dfrac{-4cos^2a\left(1-sin^2a\right)}{8sin^2a.cos^2a-8sin^2a}=\dfrac{-4cos^2a.cos^2a}{-8sin^2a\left(1-cos^2a\right)}=\dfrac{cos^4a}{2sin^2a.sin^2a}\)
\(=\dfrac{cos^4a}{2sin^4a}=\dfrac{cot^4a}{2}\)
c.
Câu c và câu b bị lặp đề
d,
\(\dfrac{1+cosa-sina}{1-cosa-sina}=\dfrac{\left(1-sina+cosa\right)^2}{\left(1-sina-cosa\right)\left(1-sina+cosa\right)}\)
\(=\dfrac{1+sin^2a+cos^2a+2cosa-2sina-2sina.cosa}{\left(1-sina\right)^2-cos^2a}\)
\(=\dfrac{2+2cosa-2sina-2sina.cosa}{1-2sina+sin^2a+sin^2a-1}=\dfrac{2\left(1+cosa\right)-2sina\left(1+cosa\right)}{2sin^2a-2sina}\)
\(=\dfrac{2\left(1-sina\right)\left(1+cosa\right)}{-2sina\left(1-sina\right)}=-\dfrac{1+cosa}{sina}=-\dfrac{1+2cos^2\dfrac{a}{2}-1}{2sin\dfrac{a}{2}cos\dfrac{a}{2}}\)
\(=-\dfrac{cos\dfrac{a}{2}}{sin\dfrac{a}{2}}=-cot\dfrac{a}{2}\)
e.
\(=\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cos2a+sin3a}{2cos3a.cos2a+cos3a}\)
\(=\dfrac{sin3a\left(2cos2a+1\right)}{cos3a\left(2cos2a+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)
f.
\(=\dfrac{cos4a.\dfrac{sin2a}{cos2a}-sin4a}{cos4a.\dfrac{cos2a}{sin2a}+sin4a}=\dfrac{cos4a.sin2a-sin4a.cos2a}{cos4a.cos2a+sin4a.sin2a}.\dfrac{sin2a}{cos2a}\)
\(=\dfrac{sin\left(2a-4a\right)}{cos\left(4a-2a\right)}.\dfrac{sin2a}{cos2a}=\dfrac{-sin2a}{cos2a}.\dfrac{sin2a}{cos2a}\)
\(=\dfrac{-sin^22a}{cos^22a}=-tan^22a\)
