Show that (cos(pi/20)+sin(pi/20))/(cos(pi/20)-sin(pi/20))=cot(pi/5) View Solution Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and ...
There are 2 steps to solve this one. Solution Share Step 1 Given that displacement function is y=14cos(15t)−17sin(15t)View the full answer Step 2 Unlock Answer UnlockPrevious question Next questionNot the question you’re looking for? Post any...
Answer to: Simplify the following. cos(frac{pi}{2} -x)tan x . A. sec x B. sin^2x C. tan x D. csc x By signing up, you'll get thousands...
百度试题 结果1 题目【题目】 \$\cos \frac { \pi } { 2 } + \sin \frac { \pi } { 4 } - \sec \frac { \pi } { 6 } + \tan \pi\$ 相关知识点: 试题来源: 解析 【解析】解: 【解析】解: 原式 _ 反馈 收藏
If the maximum possible value of (sin^(- 1)x)^2+(tan^(- 1)y)^2+pisin^(... 05:16 If the maximum value of (sec^-1 x)^2 + (cosec^-1 x)^2 approaches a, th... 15:23 (cos^(- 1)x+sec^(- 1)x+tan^(- 1)x+cot^(- 1)x)(max)+6((sin^(- 1)x)^2+pi... 08...
2cos(x)+sin(2x)=0,-180<x<=180 (deg)2sin(x)=cos(2x),0<=x<2pi (rad)sec^2(x)-tan(x)-1=0,0<=x<2pi (rad)2sin(2x)cos(3x)+cos(3x)=0,0<=x<=180 (deg) 相关知识点: 试题来源: 解析 sec^2(x)=8cos(x), -pi ...
求下列各式的值:求下列各式的值: \$\cos 0 + \sin \frac { \pi } { 2 } - \tan \pi + \sec \frac { \pi } { 3 }\$ 求下列各式的值: \$\cos 0 + \sin \frac { \pi } { 2 } - \tan \pi + \sec \frac { \pi } { 3 }\$ ...
【解析】sec*pi/30是角的弧度值sin(sinc*pi)30) 就是在这个角度情况下,单位指针长度在横方向上的投影长度8*sin(sinC+pi) 就是秒针在横方向上投影长度8*cos(sec*pi/30) 就是秒针在竖方向上投影长度答案补充:上面已经说过了。单位指针长度在横方向上的投影长度单位指针长度在纵方向上的投影长度如...
结果1 题目求下列函数的导数。求下列函数的导数。 \$y = \cos x + 2 \sec x - e ^ { x } + \sin \frac { \pi } { 2 }\$ 相关知识点: 试题来源: 解析 \$- \sin x + 2 \sec x \tan x - e ^ { x }\$ ; 反馈 收藏 ...
csc(2π−θ)= (a) sinθ (b) secθ (c) sin(−θ) (d) cosθ (c) none of these 32. Find the rectangular coordinates (x,y) of the point in polar form given by (−3,−3π) (a) (3,0) (b) (−3,0) (c) (...