Nội dung bài giảng
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:
\(\eqalign{
& \sqrt {{{\sin }^4}\alpha + {{\sin }^2}\alpha {{\cos }^2}\alpha } = \sqrt {{{\sin }^2}\alpha ({{\sin }^2}\alpha + {{\cos }^2}\alpha )} \cr
& = \sqrt {{{\sin }^2}\alpha } = |\sin \alpha | \cr} \)
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:
\(\eqalign{
& {{1 - {{\sin }^2}\alpha{{\cos }^2}\alpha} \over {{{\cos }^2}\alpha}} - {\cos ^2}\alpha\cr&= {1 \over {{{\cos }^2}\alpha }} - {\sin ^2}\alpha - {\cos ^2}\alpha \cr
& = {1 \over {{{\cos }^2}\alpha }} - 1 = {\tan ^\alpha } \cr} \)