Chứng minh hệ thức: $\rm \ \dfrac{sin^2\alpha -cos^2\alpha +cos^4\alpha}{cos^2\alpha-sin^2\alpha+sin^4\alpha}=tan^4\alpha$ 25/07/2021 Bởi Kaylee Chứng minh hệ thức: $\rm \ \dfrac{sin^2\alpha -cos^2\alpha +cos^4\alpha}{cos^2\alpha-sin^2\alpha+sin^4\alpha}=tan^4\alpha$
\[\begin{array}{l}\dfrac{{{{\sin }^2}\alpha – {{\cos }^2}\alpha + {{\cos }^4}\alpha }}{{{{\cos }^2}\alpha – {{\sin }^2}\alpha + {{\sin }^4}\alpha }} = {\tan ^4}\alpha \\VT = \dfrac{{{{\sin }^2}\alpha – {{\cos }^2}\alpha (1 – {{\cos }^2}\alpha )}}{{{{\cos }^2}\alpha – {{\sin }^2}\alpha (1 – {{\sin }^2}\alpha )}}\\ = \dfrac{{{{\sin }^2}\alpha – {{\cos }^2}\alpha {{\sin }^2}\alpha }}{{{{\cos }^2}\alpha – {{\cos }^2}\alpha {{\sin }^2}\alpha }}\\ = \dfrac{{{{\sin }^2}\alpha (1 – {{\cos }^2}\alpha )}}{{{{\cos }^2}\alpha (1 – {{\sin }^2}\alpha )}}\\ = \dfrac{{{{\sin }^4}\alpha }}{{{{\cos }^4}\alpha }} = {\tan ^4}\alpha = VP\end{array}\] Vậy hệ thức đã được chứng minh Bình luận
Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
\dfrac{{{{\sin }^2}\alpha – {{\cos }^2}\alpha + {{\cos }^4}\alpha }}{{{{\cos }^2}\alpha – {{\sin }^2}\alpha + {{\sin }^4}\alpha }} = {\tan ^4}\alpha \\
VT = \dfrac{{{{\sin }^2}\alpha – {{\cos }^2}\alpha (1 – {{\cos }^2}\alpha )}}{{{{\cos }^2}\alpha – {{\sin }^2}\alpha (1 – {{\sin }^2}\alpha )}}\\
= \dfrac{{{{\sin }^2}\alpha – {{\cos }^2}\alpha {{\sin }^2}\alpha }}{{{{\cos }^2}\alpha – {{\cos }^2}\alpha {{\sin }^2}\alpha }}\\
= \dfrac{{{{\sin }^2}\alpha (1 – {{\cos }^2}\alpha )}}{{{{\cos }^2}\alpha (1 – {{\sin }^2}\alpha )}}\\
= \dfrac{{{{\sin }^4}\alpha }}{{{{\cos }^4}\alpha }} = {\tan ^4}\alpha = VP
\end{array}\]
Vậy hệ thức đã được chứng minh