Class 12 MATHS `intsec^2xcosec^2xdx... ∫sec2xcosec2xdx Video Solution | ShareSave Answer Step by step video & image solution for intsec^2xcosec^2xdx by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams....
View Solution IfCotθ=√6, then the value ofcosec2θ+sec2θcosec2θ−sec2θis: यदिCotθ=√6,है, तोcosec2θ+sec2θcosec2θ−sec2θका मान ज्ञात करें | View Solution
integration sec^2(x)/cosec^2(x) = intg 1/cos^2(x)/1/sin^2(x) = intg tan^2(x) =intg sec^2(x)- intg dx =tanx-x+c Rishi Sharma Last Activity: 4 Years ago Dear Student, Please find below the solution to your problem. it will becomeintegral{(tan(x))^2}dx integral {(...
(a). (sec ^2θ )((cosec) θ )= sec θ ⋅ tan θsince (sec^2 θ)((cosec) θ) = ((1(cos θ))^2)( 1(sin θ))= (1(cos^2 θ))(1(sin θ))= (sin θ)(cos^2 θ)= (sin θ)(cos θ)⋅ 1(cos θ)= sec θ ⋅ tan θSo (sec^2 θ)((cosec)θ)=sec ...
sec 函数的定义域为所有不等于 π/2+kπ(k 为整数)的实数,因为在这些点上 sin(x)=0,sec(x)无定 义。 接下来,我们来看 cosec 公式。cosec 函数是正弦函数的倒数,即 cosec(x)=1/cos(x)。根据三角函数的定义,cos(x)=邻边/斜边,因此 cosec(x)=斜边/邻边。在直角三角形中,邻边是斜边与角度所对应...
【题目】cosec^2θ+sec^2θ=cosec^2θsec^2θ 相关知识点: 试题来源: 解析 【解析】L.H.S.cosec^2θ+sec^2θ =1/(sin^2θ)+1/(cos^2θ) =(cos^2θ+sin^2θ)/(sin^2θcos^2θ) =1/(sin^2θcos^2θ) =cosec^2θsec^2θL.H.S. = R.H.S.Prove. ...
Integrate the following functions with respect to x: sec ^2x, (cosec)^2x, cos ^2 12x, tan ^2x [Hint: use sec ^2x=1+tan ^2x], cot ^2x 相关知识点: 试题来源: 解析 tan x, -cot x, 12x+ 12sin x, \ tan x-x, -cot x-x 反馈 收藏 ...
cot^2θ+tan^2θ-2C. sec^2θ-cosec^2θD. cos^2θ-tan^2θ+2E. cost^2θ+tan^2θ 答案 A相关推荐 1(cosθ+tanθ)^2 等于(A. cose^2θsine^2θB. cot^2θ+tan^2θ-2C. sec^2θ-cosec^2θD. cos^2θ-tan^2θ+2E. cost^2θ+tan^2θ ...
【题目】若(cosec^2(2π-θ))/(sec^2(π+θ))=3 θ) (3π)/2θ2π 302π,求sec2(π+0)sinθ的值. 相关知识点: 试题来源: 解析 【解析】(1-sin^2θ)/(sin^2θ)-3sin^2θ-1/4,θ∈((3π)/2,2π) θ∈((3π)/2,2π)sinθ=-1/2 ...
6. (i) Use an appropriate double angle formula to show that cosec2x = λ cosec x sec x,and state the value of the constant A.(ii)Solve, for 0≤02π, the equation3sec^2θ+3secθ=2tan^2θYou must show all your working. Give your answers in terms of . 相关知识点: 试题来源...