(理)已知\alpha 为钝角,tan(\alpha +\frac{\pi}{4})=\frac{tan\alpha +1}{1-tan\alpha }=-\frac{1}{7},求: (1)tan\alpha (2)\frac{cos2\alpha +1}{\sqrt{2}cos(\alpha -\frac{\pi}{4})-sin2\alpha } (文)已知tan\alpha =2,求下列各式的值: (1)\frac{2cos^2\alpha }...
因为\tan(\alpha+\frac{\pi}{4})=\frac{\tan \alpha +1}{1-\tan \alpha }=\frac{1}{7},解得\tan\alpha =-\frac{3}{4},又因为\alpha \in (\frac{\pi}{2},\pi),因此\left\{\eqalign{& \sin\alpha=\frac{3}{5}\cr& \cos \alpha =-\frac{4}{5} }\right.,因此...
因为\tan(\alpha +{\pi\over4})={1+\tan \alpha\over 1-\tan \alpha}=2, 所以解得\tan\alpha ={1\over 3}, 因为\cos2\alpha =\cos^2\alpha -\sin^2\alpha ={\cos^2\alpha -\sin^2\alpha\over\cos^2\alpha +\sin^2\alpha} ={1-\tan ^2\alpha\over 1+\tan^2 \alpha} ={...
所以\sin \left( \alpha +\frac{ \pi }{4} \right)=\sqrt{1-{{\cos }^{2}}\left( \alpha +\frac{ \pi }{4} \right)}=\frac{2\sqrt{5}}{5}, 所以\tan \left( \alpha +\frac{ \pi }{4} \right)=\frac{\sin \left( \alpha +\frac{ \pi }{4} \right)}{\cos \left(...
百度试题 结果1 题目如果_ ,求 \$\tan \left( \alpha + \frac { \pi } { 4 } \right)\$ 相关知识点: 试题来源: 解析 解: 解: 解: 解: 反馈 收藏
Class 11 MATHS If `alpha+beta=pi/(4), then (1+"tan"alph... If α+β=π4,then(1+tanα)(1+tanβ) is A 1 B 2 C −1 D −2 Video Solution Struggling With Trigonometric F...? Get Allen’s Free Revision Notes Free ALLEN Notes Text SolutionGenerated By DoubtnutGPT The correct...
【解析】 【解析】 \$- : ( 1 - \tan \alpha ) / ( 1 + \tan \alpha ) = \tan ( \pi / 4 ) \cdot \tan \alpha / 1 + t\$ 【解析】 \$- : ( 1 - \tan \alpha ) / ( 1 + \tan \alpha ) = \tan ( \pi / 4 ) \cdot \tan \alpha / 1 + t\$ 【解析...
\$1 . \tan \left( \alpha + \frac { \pi } { 4 } \right) 0\$ 则,下列结论正确的是 \$1 . \tan \left( \alpha + \frac { \pi } { 4 } \right) 0\$ 则,下列结论正确的是 \$1 . \tan \left( \alpha + \frac { \pi } { 4 } \right) 0\$ 则,下列结论正...
$\because \tan \left(\alpha +\dfrac{\pi }{4}\right)=\dfrac{1}{7}$$\therefore \dfrac{1+\tan \alpha }{1-\tan \alpha }=\dfrac{1}{7}$解得$\tan \alpha =-\dfrac{3}{4}$,$\because \alpha \in (\dfrac{\pi }{2}$,$\pi )$,$\because \sin ^{2}\alpha +\cos ...
\sin \frac{4}{3}\pi =-\frac{\sqrt{3}}{2},\cos \frac{5}{6}\pi =-\frac{\sqrt{3}}{2},故P\left( -\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2} \right).∴\tan \alpha =1.(2)原式=\frac{-\sin \alpha \cdot \sin \alpha }{-\sin \alpha \cos \alpha }-\frac{\...