If cos(A-B)=1/2 and sin(A+B)=1/2 then find the smallest value of A and... 02:40 If cos theta + sin theta = sqrt2 cos theta " then" cos theta - sin th... 03:11 If tan theta = (5)/(4) , then the value of ((3 sin
If :tanθ+cotθ=2,then:sinθ+cosθ View Solution (secθ+tanθ)(1−sinθ)=cosθ View Solution (sinθ+cosθ)(tanθ+cotθ)= View Solution cosθ(tanθ+2)(2tanθ+1)=? cosθ(tanθ+2)(2tanθ+1)=? tanθin terms ofcosθis... Iftanθ=1, thencosθ=………. Ifcosθ.sinθ...
【解析】 ∵$$ \tan \theta = - \frac { 1 } { 3 } , $$ ∴$$ \frac { \sin \theta } { \cos \theta } = - \frac { 1 } { 3 } $$① ∵$$ \sin ^ { 2 } \theta + \cos ^ { 2 } \theta = 1 $$② 由①②得$$\left\{ \begin{matrix} \sin \theta = \...
If x = 5 tan theta, express sin theta in terms of x. What is the value of \sin \dfrac{\theta}{2} if \cos \theta = \dfrac{\sqrt{7{4} and 270^\circ \lt \theta \lt 360^\circ? Find the value of sin(2 theta) if cos(theta) = 3/5 and (theta) is in quadrant 4. ...
(说明:y=arctanx是y=tanx在-π/2≤x≤π/2的反函数) ②值域(R_{f}):\displaystyle(-\frac{\pi}{2},\frac{\pi}{2}) ③周期(T): 无 ④奇偶性:奇函数 ⑤单调性:在定义域内单增 ⑥有界性:函数在定义域内有界,即 \displaystyle-\frac{\pi}{2}< arctanx<\frac{\pi}{2}(x\in R) ...
【题目】.根据下列条件,确定θ是第几象限角.(1)cosθ与tanθ异号;(2)cosθ与tanθ同号;(3)sinθ与 $$ \cos \theta $$异号;(4)sinθ与tanθ同号. 相关知识点: 试题来源: 解析 【解析】(1)θ是第三或第四象限角; (2)θ是第一或第二象限角; (3)θ是第二或第四象限角; (4)θ是...
Answer to: Write the following in terms of sin(theta) and cos(theta); then simplify if possible. sin(theta) + cos(theta) tan(theta). By signing up,...
若$$ \tan \theta = 2 $$,则$$ \fra c { 3 \cos \theta - \sin \theta } { \cos \theta + \sin \theta } = \fra c { 1 } { 3 } . $$9. 已知$$ \sin \alpha \cos \alpha = \fra c { 1 } { 2 } $$,则$$ \sin \alpha - \cos \alpha = 0 . $$10. ...
(ii)cos2θ…...(in terms oftanθ). (iii)tan2θ…...(in terms oftanθ). View Solution Ifu=(1+cosθ)(1+cos2θ)−sinθ.sin2θ,v=sinθ(1+cos2θ)+sin2θ(1+cosθ), thenu2+v2= View Solution Exams IIT JEE NEET UP
Ifsecθ(cosθ+sinθ)=√2, then what is the value of(2sinθ)/(cosθ−sinθ)? यदिsecθ(cosθ+sinθ)=√2है,तो(2sinθ)/(cosθ−sinθ)का मान क्या है? If sec theta + tan theta = m (>1) , then the value of sin theta is...