( ((sin)(θ ))/((cos)(θ ))+((cos)(θ ))/((sin)(θ ))) 相关知识点: 试题来源: 解析 Convert from ( ((sin)(θ ))/((cos)(θ ))) to ( (tan)(θ )). ( (tan)(θ )+((cos)(θ ))/((sin)(θ ))) Convert from ( ((cos)(θ ))/((sin)(θ ))) to ( (cot)...
Step by step video & image solution for Simplify sec theta[[tan theta,sec theta],[sec theta,-tan theta]]-tan theta[[sec theta,tan theta],[tan theta,-sec theta]] by Maths experts to help you in doubts & scoring excellent marks in Class 11 exams.Updated...
Simplify, \frac{sec(x)sin(x)}{tan(x) + cot(x)}. Simplify. (t^{-4})^{-9} t^2 Simplify: x/3 - x/4. Simplify: \frac{\frac{12}{5{\frac{3}{25 Explore our homework questions and answers library Search Browse Browse by subject...
Simplify: (i) (cos theta)/(sin (90^@ + theta)) + (sin (-theta))/(sin(180^(@) + theta)) - (tan(90^(@)+theta))/(cot theta)
Answer to: Write the following in terms of sin theta and cos theta, then simplify if possible. \frac{\tan\theta}{\cot\theta} By signing up, you'll...
Question: Simplify: {eq}\displaystyle 1) \frac{-\sin(-x) \cos x \sec x \csc x \tan x}{\cot x} \\2) \frac{1 + \tan^2 \theta}{\csc^2 \theta} + \sin^2 \theta + \frac{1}{\sec^2 \theta} {/eq} Simpl...
Simplify. (cos theta + sin theta)^2 + (cos theta - sin theta)^2 Write cos 48 in terms of its cofunction. If y = integral_{1}^{x^{2 cos (t) d t, then find d y / d x a) 2 x cos(x^{2}) b) cos(x) - cos(1) c) cos(x^{2} ) d. 2 x cos(x) ...
Is there a way to simplify the proof of different vecot calculus identities, such as grad of f*g, which is expandable. And also curl of the curl of a field. Is there a more convenient way to go about proving these relations than to go through the long calculations of actually performin...
Homework Statement csc(pi/2-x)/cos(x+pi/2) + cot(pi/2-x) The Attempt at a Solution 1/cosx/-sinx + sinx/cosx -1/sinxcosx+ sinx/cosx(sinx/sinx) (sin2x - 1) / sinxcosx -cos2x/sinxcosx -cosx/sinx -cotx is this right the answer key says it is cosx but that could be wrong...
Answer to: Simplify. sin x sec x By signing up, you'll get thousands of step-by-step solutions to your homework questions. You can also ask your...