Question

Simplify the expression:
(1/1-cosx)+(1/1+cosx)

-Please help! :D

Answer 1

why not 2 csc^2(x)

Answer 2

Just like adding any other fractions together, your first step is to get a common denominator. This means that you multiply the first fraction by (1+cosx)/(1+cosx) and the second fraction by (1-cosx)/(1-cosx):

1*(1+cosx)/[(1-cosx)*(1+cosx)] + 1*(1-cosx)/[(1+cosx)*(1-cosx)]

You can simplify the denominator by multiplying (using the FOIL method). It simplifies to 1- (cosx)^2.

Simplifying the equation above you get:
(1 + cosx + 1 - cosx)/[1 - (cosx)^2]

The cosines in the numerator cancel each other out, leaving 2/[1 - (cosx)^2].

You can further simplify the denominator by using a trig identity: (sinx)^2 + (cosx)^2 = 1
Therefore, (sinx)^2 = 1 - (cosx)^2

By substituting (sinx)^2 for [1 - (cosx)^2] you get a final answer of 2/(sinx)^2

Answer 3

the answer choices include:
2cscx
csc^2x
2sec^2x
2csc^2x

Answer 4

well there you go, problem solved

Answer 5

\[\huge \frac{1}{1-\cos x} + \frac{1}{1+\cos x}\]
\[\huge \frac{1+\cos x +1-\cos x}{(1-\cos x)(1+\cos x)}\]
\[\huge \frac{2}{1-\cos^2 x}=\huge \frac{2}{\sin^2 x} = 2cosec^2x\]

@RavenclawUnited

Answer 6

\[\huge 2cosec^2x = 2\csc^2 x\]

Answer 7

how did you get to 2 csc^2x @campbell_st

Answer 8

ohh okay

Answer 9

\[\huge \frac{1}{sinx} = \csc x \rightarrow \frac{1}{\sin^2x} = \csc^2 x\]
@RavenclawUnited

Answer 10

well its simply

2(1/sin(x))^2 and 1/sin = csc

so substituting

2(csc(x))^2

which is written

2 csc^2(x)

Answer 11

Thank you @campbell_st and @dpasingh