The general solution of tan(3theta)=1 is 01:50 The general solution of tan theta tan 2theta=1 is 05:48 If tan(2theta)=cottheta implies theta=(ninz) 04:53 The solution set of sin 3 theta = (sqrt(10 + 2sqrt(5)))/(4) is 02:21 If sin(theta-25^(@))=-(sqrt(3))/(2)...
(2cos^(2)theta-1)/(sinthetacostheta)=cottheta-tantheta. 02:30 (1+cos^(2)A)/(sin^(2)A)=2cosec^(2)A-1. 02:06 निम्नलिखित सर्वसमिकाओ को सिद्ध कीजिए:(1)/(secA+tanA)-(1)/(cosA)=(1)/...
∵tan\theta =2=\frac{sin\theta }{cos\theta } ∴sin\theta =2cos\theta tan^2\theta =\frac{1-cos^2\theta }{cos^2\theta }=\frac{1}{cos^2\theta }-1=4 ∴cos^2\theta =\frac{1}{tan^2\theta +1}=\frac{1}{5} ∴1+sin\theta cos\theta =1+2cos^2\theta =1+\frac{2}...
A. 2 B. \frac{1}{2} C. -\frac{1}{2} D. -2 相关知识点: 试题来源: 解析 A ∵\tan\theta =2, ∴\tan(\theta +\pi) =\frac{\tan \theta +\tan \pi}{1-\tan \theta \tan \pi} =\frac{\tan \theta }{1} =2, 故本题的正确答案为A。
结果1 题目已知\( \tan \theta = \frac{1}{2} \),则\( \sin \theta \)的值为: A. \( \frac{\sqrt{5}}{5} \) B. \( \frac{\sqrt{10}}{10} \) C. \( \frac{2\sqrt{5}}{5} \) D. \( \frac{\sqrt{10}}{5} \) ...
解析 \frac{60}{11} 解:\tan\theta =1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+……+\frac{1}{5}-\frac{1}{6} =1-\frac{1}{6} =\frac{5}{6} \tan2\theta =\frac{2\tan\theta }{1-\tan^2\theta }=\frac{\frac{5}{3}}{1-\frac{25}{36}}=\frac{60}{11}...
{2}\cdot\dfrac{2\sin\theta\cos\theta +{\cos}^2 \theta -{\sin}^2 \theta}{{\sin}^2 \theta +{\cos}^2 \theta} \\ & =\dfrac{\sqrt{2}}{2}\cdot\dfrac{2\tan{\theta}+1-{\tan}^2 \theta}{1+{\tan}^2 \theta} \\ & =\dfrac{7\sqrt{2}}{10}.\end{split}\e...
\theta -2{{\cos }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta +\sin \theta \cos \theta -2{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }=\dfrac{{{\tan }^{2}}\theta +\tan \theta -2}{{{\tan }^{2}}\theta +1}=\dfrac{4+2-2}{4+...
【正确解法】由\(\tan \theta =1\),得\(\theta =k \pi +\dfrac{ \pi }{4}(k \in \mathbf{Z})\),故\(\sin (2 \theta + \varphi )= \sin \left(\dfrac{ \pi }{2}+ \varphi \right)= \cos \varphi\),\(\because \sin (2 \theta + \varphi )=3 \sin \varphi\),\...
If sin {eq}\theta {/eq} > 0 and tan {eq}\theta {/eq} = -{eq}\frac{1}{2} {/eq}, find the exact value of cos {eq}\theta {/eq}. Cosine of an Angle: Suppose we are given the tangent of an angle and we are required to find the cosine of ...