This qustion is from a pre calculus class.

Question

Can you help me with this problem:

 

Verify the identity: sinx/1+cosx  +1+cosx/sinx= 2cscx

Answer 1

I believe you are missing some parentheses, but assuming my fractions, here is the solution:

 

• sinx/(1+cosx) + (1+cosx)/sinx (we want to start with this side, because it's more complex, and will hopefully narrow down easily
• (sin²x)/[(1+cosx)(sinx)] + (1+2cosx+cos²x)/[(1+cosx)(sinx)]   ~   This is just getting a common denominator
• (sin²x+cos²x+1+2cosx)/[(1+cosx)sinx]   ~ Regrouping...note that first part is just 1 (pythagorean identity)
• (2+2cosx)/[(1+cosx)sinx]
• 2/sinx   ~  (1+cosx) cancels
• 2cscx   ~  1/sinx = csc x

 

Wow, that was harder to type than it was to solve.  Hope you can make sense of this.  It's MUCH easier to see using the actual fractions, rather than trying to type them.

Answer 2

sin x/(1+cos x) + (1+cos x)/sin x = 2 csc x

 

Put the fractions on the LHS of the equation over a common denominator, (sin x)(1 + cos x):

 

{ (sin x)(sin x) + (1 + cos x)(1 + cos x) } / (sin x)(1 + cos x)

 

= {sin² x + 1 + 2 cos x + cos² x} / (sin x)(1 + cos x)

 

sin2 x + cos² x = 1, so:

 

= {2 + 2 cos x} / (sin x)(1 + cos x)

 

= 2(1 + cos x) / (sin x)(1 + cos x)

 

= 2/sin x

 

= 2 csc x