How do you verify #sin x + cos x * cot x = csc x#?

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Answer 1 · 2016-03-02T01:43:37

Recall the following reciprocal, tangent, and Pythagorean identities:

#1#. #color(orange)cotx=1/tanx#

#2#. #color(blue)cscx=1/sinx#

#3#. #color(purple)tanx=sinx/cosx#

#4#. #color(brown)(sin^2x+cos^2x)=1#

Proving the Identity
#1#. Simplify by rewriting the left side of the identity in terms of #sinx# and #cosx#.

#sinx+cosx*color(orange)cotx=color(blue)cscx#

Left side:

#sinx+cosx*1/color(purple)tanx#

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

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

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

#=sinx+cosx*(cosx/sinx)#

#2#. Simplify by multiplying #cosx# by #cosx/sinx#.

#=sinx+cos^2x/sinx#

#3#. Find the L.C.M. (lowest common multiple) to rewrite the left hand side as a fraction.

#=(sinx(sinx)+cos^2x)/sinx#

#4#. Simplify.

#=color(brown)((sin^2x+cos^2x))/sinx#

#=1/sinx#

#=color(green)(cscx)#

#:.#, left side#=#right side.