Đáp án:
\[B = \tan x\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin x = \sin \left( {\pi - x} \right)\\
\sin x = - \sin \left( { - x} \right)\\
\cos x = - \cos \left( {\pi - x} \right)\\
\cos x = \cos \left( { - x} \right)\\
B = 2\cos x - 3\cos \left( {\pi - x} \right) + 5\sin \left( {\dfrac{{7\pi }}{2} - x} \right) + \cot \left( {\dfrac{{3\pi }}{2} - x} \right)\\
= 2\cos x + 3\cos x + 5\sin \left[ {4\pi - \left( {x + \dfrac{\pi }{2}} \right)} \right] + \cot \left( {\pi + \left( {\dfrac{\pi }{2} - x} \right)} \right)\\
= 5\cos x + 5\sin \left( { - \left( {x + \dfrac{\pi }{2}} \right)} \right) + \cot \left( {\dfrac{\pi }{2} - x} \right)\\
= 5\cos x - 5\sin \left( {x + \dfrac{\pi }{2}} \right) + \dfrac{{\cos \left( {\dfrac{\pi }{2} - x} \right)}}{{\sin \left( {\dfrac{\pi }{2} - x} \right)}}\\
= 5\cos x - 5\cos \left( {\dfrac{\pi }{2} - \left( {x + \dfrac{\pi }{2}} \right)} \right) + \dfrac{{\sin x}}{{\cos x}}\\
= 5\cos x - 5\cos \left( { - x} \right) + \tan x\\
= 5\cos x - 5\cos x + \tan x\\
= \tan x
\end{array}\)
