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Đơn giản các biểu thức
a) \(\sqrt {{{\sin }^4}\alpha + {{\sin }^2}\alpha {{\cos }^2}\alpha } \)
b) \({{1 – \cos \alpha } \over {{{\sin }^2}\alpha }} – {1 \over {1 + \cos \alpha }}\,\,(\sin \alpha \ne 0)\)
c) \({{1 – {{\sin }^2}\alpha {{\cos }^2}\alpha } \over {{{\cos }^2}\alpha }} – {\cos ^2}\alpha \,\,\,(cos\alpha \ne 0)\)
Đáp án
a) Ta có:
\(\eqalign{
& \sqrt {{{\sin }^4}\alpha + {{\sin }^2}\alpha {{\cos }^2}\alpha } = \sqrt {{{\sin }^2}\alpha ({{\sin }^2}\alpha + {{\cos }^2}\alpha )} \cr
& = \sqrt {{{\sin }^2}\alpha } = |\sin \alpha | \cr} \)
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b) Ta có:
\(\eqalign{
& {{1 – \cos \alpha } \over {{{\sin }^2}\alpha }} – {1 \over {1 + \cos \alpha }}= {{1 – \cos \alpha } \over {1 – {{\cos }^2}\alpha }} – {1 \over {1 + \cos \alpha }} \cr
& = {1 \over {1 + \cos \alpha }} – {1 \over {1 + \cos \alpha }} = 0 \cr} \)
c) Ta có:
\(\eqalign{
& {{1 – {{\sin }^2}\alpha{{\cos }^2}\alpha} \over {{{\cos }^2}\alpha}} – {\cos ^2}\alpha\cr&= {1 \over {{{\cos }^2}\alpha }} – {\sin ^2}\alpha – {\cos ^2}\alpha \cr
& = {1 \over {{{\cos }^2}\alpha }} – 1 = {\tan ^\alpha } \cr} \)