Question

[Buddy]

Rút gọn các biểu thức sau:

a)    \(\frac{1}{{\tan \alpha  + 1}} + \frac{1}{{\cot \alpha  + 1}}\)

b)    \(\cos \left( {\frac{\pi }{2} - \alpha } \right) - \sin \left( {\pi  + \alpha } \right)\)

c)    \(\sin \left( {\alpha  - \frac{\pi }{2}} \right) + \cos \left( { - \alpha  + 6\pi } \right) - \tan \left( {\alpha  + \pi } \right)\cot \left( {3\pi  - \alpha } \right)\)

Answer 1

[Hà Quang Minh]

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

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

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