Question #e026e

2 Answers
Narad T.
Aug 17, 2017

See the proof below

Explanation:

We need

#cscx=1/sinx#

#cotx=cosx/sinx#

#cos^2x+sin^2x=1#

Therefore

#LHS =(1/(cscx-cotx))-(1/(cscx+cotx))#

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

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

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

#=sinx/sin^2x*2cosx#

#=2/(sinx/cosx)#

#=2/tanx#

#=RHS#

#QED#

Anjali G
Aug 17, 2017

#(1/(cscx - cotx)) -(1/(cscx + cotx)) = 2/tanx#

Multiply the fractions on the left side by the conjugates of the denominators (which is the same as multiplying by one):

#"Left side:"#
#frac{1}{cscx-cotx} color(blue)(*frac{cscx+cotx}{cscx+cotx}) - frac{1}{cscx + cotx} color(blue)(*frac{cscx-cotx}{cscx-cotx})#

# "LS"= frac{(cscx+cotx) - (cscx - cotx)}{csc^2 x -cot^2 x}#

Using the pythagorean trig identity, we know that #csc^2x - cot^2x = 1#
#"LS" = frac{color(red)(cscx) + cotx color(red)(-cscx) + cotx}{color(blue)((csc^2x - cot^2x))}#

#"LS" = 2cotx#

#"LS" = 2/tanx#

#="Right side"#

#QED#

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