Đáp án:
$6)
a) A=\tan\alpha\\
b) B=8(1+\cos\alpha)\\
c)-\cot\dfrac{\alpha}{2}$
Giải thích các bước giải:
$6)
a) A=\dfrac{\sin2\alpha+\sin\alpha}{1+\cos2\alpha +\cos\alpha}\\
=\dfrac{2\sin\alpha\cos\alpha+\sin\alpha}{\sin^2\alpha+\cos^2\alpha+\cos^2\alpha-\sin^2\alpha+\cos\alpha}\\
=\dfrac{\sin\alpha(2\cos\alpha+1)}{2\cos^2\alpha+\cos\alpha}\\
=\dfrac{\sin\alpha(2\cos\alpha+1)}{\cos\alpha(2\cos\alpha+1)}\\
=\tan\alpha\\
b) B=\dfrac{4\sin^2\alpha}{1-\cos^2\dfrac{\alpha}{2}}\\
=\dfrac{4(1-\cos^2\alpha)}{\sin^2\dfrac{\alpha}{2}}\\
=\dfrac{4(1-\cos\alpha)(1+\cos\alpha)}{\dfrac{1-\cos\alpha}{2}}\\
=8(1+\cos\alpha)\\
c)\dfrac{1+\cos\alpha-\sin\alpha}{1-\cos\alpha-\sin\alpha}\\
=\dfrac{1+2\cos^2\dfrac{\alpha}{2}-1-2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}}{1-1+2\sin^2\dfrac{\alpha}{2}-2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}}\\
=\dfrac{2\cos^2\dfrac{\alpha}{2}-2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}}{2\sin^2\dfrac{\alpha}{2}-2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}}\\
=\dfrac{2\cos\dfrac{\alpha}{2}(\cos\dfrac{\alpha}{2}-\sin\dfrac{\alpha}{2})}{2\sin\dfrac{\alpha}{2}(\sin\dfrac{\alpha}{2}-\cos\dfrac{\alpha}{2})}\\
=\dfrac{2\cos\dfrac{\alpha}{2}(\cos\dfrac{\alpha}{2}-\sin\dfrac{\alpha}{2})}{-2\sin\dfrac{\alpha}{2}(-\sin\dfrac{\alpha}{2}+\cos\dfrac{\alpha}{2})}\\
=-\cot\dfrac{\alpha}{2}$
