Question

(1/cosx+1)+(1/cosx-1)=-2cscxcotx

Answer 1

\[ \frac{1}{\cos x + 1} + \frac{1}{\cos x - 1} = -2\csc x\;\cot x\]
\[ \frac{(\cos x - 1) + (\cos x + 1)}{(\cos x - 1)(\cos x + 1)} = -2\csc x \cot x\]
\[ \frac{(2\cos x)}{(\cos^2 x - 1)} = -2\csc x \cot x\]
\[ \frac{(2\cos x)}{(-\sin^2 x)} = -2\csc x \cot x\]
\[ -2 \csc x \cot x = -2\csc x \cot x\]

Answer 2

can somebody please explain the last step?

Answer 3

\[-2(1/sinx)(\cos/\sin) =-2 (cscx)(cotx)\]

Answer 4

(2cosx)(−sin2x), i mean this part. how does he go from there to -2csc

Answer 5

split (2 cosx )/(sin^2x)

Answer 6

so basically it becomes (cosx/-sin) and (1/-sin)?

Answer 7

\[-2 \frac{cosx}{\sin^{2}}= -2 * \frac{1}{sinx} *\frac{\cos}{sinx} =-2 cscx*cotx\]

Answer 8

ah i see. thank you for the explaination

Answer 9

u r welcome :)

Answer 10

so the whole number doesn't mean there are 2 cosines. there can only be more than one cosine depending on the exponent. that part confused me a little. thank you.