Đáp án:
\(\begin{array}{l}
a,\\
A = 0\\
b,\\
B = 2
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin x = \cos \left( {90^\circ - x} \right)\\
\cos x = \sin \left( {90^\circ - x} \right)\\
\tan x = \cot \left( {90^\circ - x} \right)\\
\cot x = \tan \left( {90^\circ - x} \right)\\
\tan x.\cot x = 1\\
{\sin ^2}x + {\cos ^2}x = 1\\
a,\\
A = \dfrac{{\tan 78^\circ }}{{\cot 12^\circ }} - {\sin ^2}54^\circ - {\sin ^2}36^\circ \\
= \dfrac{{\tan 78^\circ }}{{\tan \left( {90^\circ - 12^\circ } \right)}} - {\sin ^2}54^\circ - {\cos ^2}\left( {90^\circ - 36^\circ } \right)\\
= \dfrac{{\tan 78^\circ }}{{\tan 78^\circ }} - {\sin ^2}54^\circ - {\cos ^2}54^\circ \\
= 1 - \left( {{{\sin }^2}54^\circ + {{\cos }^2}54^\circ } \right)\\
= 1 - 1\\
= 0\\
b,\\
B = \tan 46^\circ .\tan 44^\circ + \dfrac{{1 - {{\sin }^2}42^\circ }}{{{{\sin }^2}48^\circ }}\\
= \tan 46^\circ .\cot \left( {90^\circ - 44^\circ } \right) + \dfrac{{\cos 42^\circ }}{{{{\cos }^2}\left( {90^\circ - 48^\circ } \right)}}\\
= \tan 46^\circ .\cot 46^\circ + \dfrac{{{{\cos }^2}42^\circ }}{{{{\cos }^2}42^\circ }}\\
= 1 + 1\\
= 2
\end{array}\)
