I need help proving this identity
1/cosx - cosx = sinxtanx
I have helped you with two of these now.
I think it is time for you to try them.
Hint:
on the LS, find a common denominator and add the two terms
Show me your steps and I will help you from there.
LS = 1/cosx - cosx/cosx
= 1-cosx/cosx
= sinx/cosx
I know this answer is extremely wrong. I have no idea what to do.
Reiny, are you still here. I need help quick. I'm really bad at trigonometric identities. I need help.
Ls = 1/cosx - cos^2 x/cosx
= (1-cos^2 x)/cox
= sin^2 x/cosx
= sinx sinx/cox
= sinx tanx
= RS
How can you change
1/cosx - cosx to 1/cosx - cosx/cosx ??
you can't just toss in an extra cosx at the bottom
that last term of cosx/cosx would reduce to 1
Have a tough time understanding calculus , in need of help please, appreciate you in advance <3
(i) Find the general solution of the Differential equation
𝑑2𝑦/𝑑𝑡2 = −4𝑡2 + 5𝑡 + 3
(ii)Hence, find the solution when 𝑡 = 0 , 𝑦 = 1 and 𝒅𝒚/𝒅𝒕= 1.
To prove the given identity, we need to manipulate the left side of the equation until it matches the right side.
Starting with the left side, perform the following steps:
Step 1: Identify the least common denominator (LCD) of the fractions involved. In this case, the LCD is cosx.
Step 2: Write both fractions over the common denominator:
1/cosx - cosx = (1 - cos^2x) / cosx
Step 3: Since 1 - cos^2x can be factored further, rewrite the numerator:
(1 - cos^2x) = sin^2x
After these steps, the equation becomes:
sin^2x / cosx
Step 4: Recall the trigonometric identity for tangent:
tanx = sinx / cosx
Step 5: Divide sin^2x by cosx, resulting in:
sinx * sinx / cosx = sinxtanx
By following these steps, the left side (1/cosx - cosx) is successfully transformed into the right side (sinxtanx), proving the identity.