How do I simplify sin^2(x) /1-cos(x)?
[math]\dfrac{\sin^2 x}{1-\cos x}=\dfrac{1-\cos^2x}{1-\cos x} \tag*{}[/math][math]\dfrac{1-\cos^2x}{1-\cos x}=\dfrac{(1-\cos x)(1+\cos x)}{1-\cos x} \tag*{}[/math][math]\dfrac{(1-\cos x)(1+\cos x)}{1-\cos x}=1+\cos x \tag*{}[/math]Tools used:[math]\sin^2x+\cos^2x\equiv 1\tag*{}[/math][math]1-a^2\equiv (1+a)(1-a)\tag*{}[/math]
Simplify (1-cos theta)(1+cos theta)/(1-sin theta)(1+sin theta) math question?
(1-cos theta)(1+cos theta)/(1-sin theta)(1+sin theta)= (1 - cos^2(theta) ) / (1 -sin^2(theta) ) sin^2(theta) = 1- cos^2(theta) cos^2(theta) = 1 - sin^2(theta) sin^2(theta) / cos^2(theta) = tan^2(theta) ==== Final answer tan^2(theta)
How do I simplify 2cos theta-cos theta (1-3tan theta)?
Suppose. theta=x2.cos x-cos x.(1–3tan x)=?=cos x.(2–1+3tan x)=cos x.(1+3.tan x)= cos x.(1+3.sin x/cos x)=cos x.(cos x+3.sin x)/cos x= cos x + 3.sin x. Answer.
Simplify cos(theta)tan^2(theta)+cos(the...
cosθ * tan^2θ + cosθ = cos * sin^2θ / cos^2θ + cosθ = sin^2θ / cosθ + cos^2θ / cosθ = (sin^2θ + cos^2θ) / cosθ = 1 / cosθ = secθ Ta da.
How to simplify: (sinΘ / cscΘ) + (cosΘ / secΘ)?
im not gonna use theta cause they are annoying, pretend they are there csc = 1/sin , sec = 1/cos so this would give [(sin)/(1/sin)] = [(cos) / (1/cos)] Since these are complex fractions, you need to multiply the top by the reciprocal which would give u sin*sin + cos*cos This gives sin^2+cos^2 and this = 1
How can I simplify (cos(x) / (1-sin(x))) + (cos(x) /1+sin(x)) to 2sec(x)?
Here we make the denominators equal…
PLEASE HELP! simplify 1-cos^2 theta?
Could someone please explain to me how you would simplify 1-cos^2theta using the identity sin^2theta+cos^2theta=1? I've tried to figure it out and it's just not clicking because I'm not sure how to do it. If you could explain why you would do each step or at least try to it would be a big help in me understanding it. However, any ounce of help will be appreciated thanks!
How you simplify (1/sin x + cos x/sin x) (1-cos x)?
(1/sin x + cos x/sin x) (1-cos x) = (1+cos x/sin x) ( 1-cos x) = ((1+cos x)(1-cos x))/sin x= [1-(cos x)^2]/sin x = (sin x)^2 / sin x = sin x
How do you show that sec theta - 1 ÷ sec theta + 1 = sin theta ÷ 1 + cos theta whole square?
How do you show that sec theta - 1 ÷ sec theta + 1 = sin theta ÷ 1 + cos theta whole square?Quora has a “math mode” which makes typing and displaying mathematical notation much cleaner and potentially less ambiguous. You don’t have to write things out in words, and you can get proper grouping. I use An introduction to beautiful math on Quora as a reference to most math formatting I want to do on Quora. It doesn’t cover everything, but it’s a decent introduction.I don’t know, without guessing, if you want to prove:[math]\sec\theta - \frac{1}{\sec\theta} + 1 = \left(\frac{\sin\theta}{1} + \cos\theta\right)^2[/math]or if you want to prove:[math]\frac{\sec\theta - 1}{\sec\theta + 1} = \left(\frac{\sin\theta}{1+\cos\theta}\right)^2[/math]Assuming the second, the first thing I’d do is simplify the LHS to convert all the trig functions into sines and cosines (like the RHS):[math]\begin{align} \frac{\sec\theta -1}{\sec\theta + 1} & = \frac{\frac{1}{\cos\theta}-1}{\frac{1}{\cos\theta} + 1} \\ & = \frac{1 - \cos\theta}{1 + \cos\theta} \end{align}[/math]I’ll simplify the denominator by multiplying that by [math]1=\frac{1+\cos\theta}{1+\cos\theta}[/math] to get [math]\frac{1-\cos\theta}{1+\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta} = \frac{1-\cos^2\theta}{(1+\cos\theta)^2}[/math].Using the identity that [math]\sin^2\theta + \cos^2\theta = 1[/math], you get [math]\sin^2\theta = 1 - \cos^2\theta[/math], which is the numerator of the above. This gives us [math]\frac{\sin^2\theta}{(1+\cos\theta)^2} = \left(\frac{\sin\theta}{1+\cos\theta)}\right)^2[/math] as desired.In one big chain of equalities:[math]\begin{align} \frac{\sec\theta - 1}{\sec\theta + 1} & = \frac{\frac{1}{\cos\theta} - 1}{\frac{1}{\cos\theta} + 1} \\ &= \frac{\frac{1}{\cos\theta} -1}{\frac{1}{\cos\theta} + 1} \cdot \frac{\cos\theta}{\cos\theta} \\ &= \frac{1 - \cos\theta}{1 + \cos\theta} \\ &= \frac{1-\cos\theta}{1+\cos\theta} \cdot \frac{1+\cos\theta}{1+\cos\theta} \\ &= \frac{(1-\cos\theta)(1+\cos\theta)}{(1+\cos\theta)(1+\cos\theta)} \\ &= \frac{1-\cos^2\theta}{(1+\cos\theta)^2} \\ &= \frac{\sin^2\theta}{(1+\cos\theta)^2} \\ &= \left(\frac{\sin\theta}{1+\cos\theta}\right)^2 \end{align}[/math]