1/ sin bình ampha – cos bình ampha + 1 +tan bình ampha/1-tan bình ampha 13/09/2021 Bởi Lyla 1/ sin bình ampha – cos bình ampha + 1 +tan bình ampha/1-tan bình ampha
\[\begin{array}{l} \frac{1}{{{{\sin }^2} – {{\cos }^2}}} + \frac{{1 + {{\tan }^2}}}{{1 – {{\tan }^2}}}\\ = \frac{1}{{{{\sin }^2} – {{\cos }^2}}} + \frac{{1 + \frac{{{{\sin }^2}}}{{{{\cos }^2}}}}}{{1 – \frac{{{{\sin }^2}}}{{{{\cos }^2}}}}} = \frac{1}{{{{\sin }^2} – {{\cos }^2}}} + \frac{{{{\cos }^2} + {{\sin }^2}}}{{{{\cos }^2} – {{\sin }^2}}}\\ = \frac{1}{{{{\sin }^2} – {{\cos }^2}}} + \frac{1}{{{{\cos }^2} – {{\sin }^2}}} = 0 \end{array}\] Bình luận
`1/(sin^2 a-cos^2a)+(1+tan^2 a)/(1-tan^2a)` `=1/(sin^2 a-cos^2a)+(1+((sina)/(cosa))^2)/(1-((sina)/(cosa))^2)` `=1/(sin^2 a-cos^2a)+((cos^2a+sin^2a)/(cos^2a))/((cos^2a-sin^2a)/(cos^2a))` `=1/(sin^2 a-cos^2a)+1/(cos^2a) . (cos^2a)/(cos^2a-sin^2a)` `=1/(sin^2 a-cos^2a)+1/(cos^2a-sin^2a)` `=0` Bình luận
\[\begin{array}{l}
\frac{1}{{{{\sin }^2} – {{\cos }^2}}} + \frac{{1 + {{\tan }^2}}}{{1 – {{\tan }^2}}}\\
= \frac{1}{{{{\sin }^2} – {{\cos }^2}}} + \frac{{1 + \frac{{{{\sin }^2}}}{{{{\cos }^2}}}}}{{1 – \frac{{{{\sin }^2}}}{{{{\cos }^2}}}}} = \frac{1}{{{{\sin }^2} – {{\cos }^2}}} + \frac{{{{\cos }^2} + {{\sin }^2}}}{{{{\cos }^2} – {{\sin }^2}}}\\
= \frac{1}{{{{\sin }^2} – {{\cos }^2}}} + \frac{1}{{{{\cos }^2} – {{\sin }^2}}} = 0
\end{array}\]
`1/(sin^2 a-cos^2a)+(1+tan^2 a)/(1-tan^2a)`
`=1/(sin^2 a-cos^2a)+(1+((sina)/(cosa))^2)/(1-((sina)/(cosa))^2)`
`=1/(sin^2 a-cos^2a)+((cos^2a+sin^2a)/(cos^2a))/((cos^2a-sin^2a)/(cos^2a))`
`=1/(sin^2 a-cos^2a)+1/(cos^2a) . (cos^2a)/(cos^2a-sin^2a)`
`=1/(sin^2 a-cos^2a)+1/(cos^2a-sin^2a)`
`=0`