How do I prove that #cscx-sinx=cosxcotx#?

3 Answers
Monzur R.
May 19, 2017

#cscx-sinx#

#=1/sinx-sinx#

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

#=cos^2x/sinx#

#=cosx/sinx xxcosx#

#=cotxcosx#

#thereforecscx-sinx=cotxcosx# #sf(QED)#

Narad T.
May 19, 2017

See proof below

Explanation:

We need

#cscx=1/sinx#

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

#cotx=cosx/sinx#

Therefore,

#LHS=cscx-sinx#

#=1/sinx-sinx#

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

#=cos^2x/sinx#

#=cosx*cosx/sinx#

#=cosxcotx#

#=RHS#

#QED#

Tazwar Sikder
May 19, 2017

We have: #csc(x) - sin(x)#

Let's apply a standard trigonometric identity; #csc(x) = frac(1)(sin(x))#:

#= frac(1)(sin(x)) - sin(x)#

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

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

One of the Pythagorean identities is #cos^(2)(x) + sin^(2)(x) = 1#

We can rearrange it to get:

#Rightarrow cos^(2)(x) = 1 - sin^(2)(x)#

Let's apply this rearranged identity to our proof:

#= frac(cos^(2)(x))(sin(x))#

#= cos(x) times frac(cos(x))(sin(x))#

Finally, let's apply another standard trigonometric identity; #cot(x) = frac(cos(x))(sin(x))#:

#= cos(x) times cot(x)#

#= cos(x) cot(x)# #\# Q.E.D.

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