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.