Prove that #(cscx)/(1+cscx)-(cscx)/(1-cscx)=2sec^2x#?

4 Answers
Apr 8, 2017

Answer:

You need to know that #csc x = 1/sin x#, that #sec x = 1/cos x#, and the formula #sin^2 x + cos^2 x =1#

Explanation:

I assume you meant:
#(csc x) / (1 + csc x) - (csc x) / (1 - csc x) = 2 sec^2 x#

These cosecant (#csc#) everywhere are suspicious,
so let's change them since we know:
#csc x = 1/sin x#

Let's transform our left-hand-side of the equation:
#(csc x) / (1 + csc x) - (csc x) / (1 - csc x) #

Let's first put everything under the same denominator:
#= ( csc x*(1-csc x) - csc x * (1+csc x ))/( (1+csc x) (1-csc x) )#

and simplifying a bit, we get:
#= ( csc x-csc^2 x - csc x - csc^2 x ) / (1 - csc^2 x) #

so
#= (-2csc^2 x)/(1-csc^2 x)#

This then becomes
#= (-2/sin^2 x)/(1-1/sin^2 x)#

Again simplifying, we get:
#= (-2/sin^2 x)/( (sin^2 x -1)/sin^2 x )#

We cancel the #1/sin^2 x# from the numerator and denominator and we get:
#=-2/(sin^2x-1)#

Remember that there is this wonderful formula #sin^2 x + cos^2 x = 1#. Then our expression becomes:
#=-2/(sin^2x-sin^2 x-cos^2 x)#
#=2/cos^2 x#
and since #cos x = 1/sec x#, we have

#=2sec^2 x#

which is exactly the right-hand-side of the equation.

Q.E.D.

Apr 8, 2017

Answer:

Please see below.

Explanation:

#cscx/(1+cscx)-cscx/(1-cscx)#

=#(1/sinx)/(1+1/sinx)-(1/sinx)/(1-1/sinx)#

Multiplying each term by #sinx# we get

#1/(sinx+1)-1/(sinx-1)#

= #1/(1+sinx)+1/(1-sinx)#

= #(1-sinx+1+sinx)/(1-sin^2x)#

= #2/cos^2x#

= #2sec^2x#

Apr 8, 2017

Answer:

Proved in the explanation.

Explanation:

Prove: #(csc(x)) / (1 + csc(x)) - (csc(x)) / (1 - csc(x)) = 2 sec^2(x)#

Multiply the left side of the equation by 1 in the form, #sin(x)/sin(x)#

#sin(x)/sin(x)((csc(x)) / (1 + csc(x)) - (csc(x)) / (1 - csc(x))) = 2 sec^2(x)#

Please observe that, when the sine function is distributed, everywhere there was #csc(x)# becomes 1 and everywhere there was 1 becomes #sin(x)#

#1 / (sin(x) + 1) - 1/ (sin(x) - 1) = 2 sec^2(x)#

Multiply the first term by 1 in the form #(sin(x) - 1)/(sin(x) - 1)#:

#1 / (sin(x) + 1)(sin(x) - 1)/(sin(x) - 1) - 1/ (sin(x) - 1) = 2 sec^2(x)#

This makes the denominator become the difference of two squares:

#(sin(x) - 1)/(sin^2(x) - 1) - 1/ (sin(x) - 1) = 2 sec^2(x)#

Multiply the second term by 1 in the form #(sin(x) + 1)/(sin(x) + 1)#:

#(sin(x) - 1)/(sin^2(x) - 1) - 1/ (sin(x) - 1)(sin(x) + 1)/(sin(x) + 1) = 2 sec^2(x)#

This makes the denominator become the same difference of two squares:

#(sin(x) - 1)/(sin^2(x) - 1) - (sin(x) + 1)/ (sin^2(x) - 1) = 2 sec^2(x)#

Put both numerators over the common denominator:

#(sin(x) - 1 - sin(x) - 1)/ (sin^2(x) - 1) = 2 sec^2(x)#

Combine like terms:

#-2/ (sin^2(x) - 1) = 2 sec^2(x)#

Substitute #-cos^2(x)# for #(sin^2(x) - 1)#:

#-2/-cos^2(x) = 2 sec^2(x)#

Because #1/cos(x) = sec(x)#, the left side becomes the same as the right:

#2sec^2(x) = 2 sec^2(x)#

Q.E.D.

Apr 8, 2017

Answer:

Make a common denominator.

Explanation:

#(csc x) / (1 + cscx)# - #(csc x)/(1-csc x)# = #2 sec^2 x#

Taking the left hand side & multiplying to get a common denominator,

#((csc x)(1-csc x) - (csc x)(1+cscx)) / ((1+csc x)(1-cscx))#

= #(csc x - csc^2 x - csc x - csc^2 x) / (1 - csc x + csc x - csc^2 x)#

= #(-2 csc^2 x)/(1 - csc^2 x)#

= #-2/sin^2 x# / #1 - 1/sin^2 x#

( / is the divide symbol, easier to see when not in fraction form)

= #-2/sin^2 x# / #(sin^x - 1)/sin^2 x#

(Common denominator again by multiplying 1 by #sin^2 x#)

= #-2/sin^2 x# * #sin^2x / (sin^2 x - 1)#

(switch numerator & denominator and cancel out common terms)

= #-2/(sin^2 x-1)#

= #-2/-cos^2 x#

(#sin^2 x + cos ^2 x = 1#) Hence, #sin^2 x - 1 = -cos^2 x #

= #2/cos^2 x#

= #2 sec^2 x#