How do you express #tan theta - cot theta +sintheta # in terms of #cos theta #?

1 Answer
Karthik G
Jan 9, 2016

#=(-cos^3(theta)-2cos^2(theta)+cos(theta)+1)/(cos(theta)sqrt(1-cos^2theta))#

Explanation:

#tan(theta)-cot(theta)+sin(theta)#
We have to write in terms of #cos(theta)#

#color(blue)"Let us start by using the identity"#
#tan(theta) = sin(theta)/cos(theta)# and #cot(theta) = cos(theta)/sin(theta)#

We get

#tan(theta)-cot(theta)+sin(theta)#
#=sin(theta)/cos(theta) - cos(theta)/sin(theta) + sin(theta)#

#color(blue)"In order to simplify we need to use Least Common Denominator for all the fractions"#

#=(sin(theta)sin(theta))/(cos(theta)sin(theta)) -(cos(theta)cos(theta))/(cos(theta)sin(theta)) + (sin(theta)cos(theta)sin(theta))/(cos(theta)sin(theta))#

#=(sin^2(theta)-cos^2(theta) + sin^2(theta)cos(theta))/(cos(theta)sin(theta))#

#=(1-cos^2(theta)-cos^2(theta)+(1-cos^2(theta))cos(theta))/(cos(theta)sin(theta))#

#=((1-2cos^2(theta))+cos(theta)-cos^3(theta))/(cos(theta)sin(theta))#

#=(-cos^3(theta)-2cos^2(theta)+cos(theta)+1)/(cos(theta)sin(theta))#

#=(-cos^3(theta)-2cos^2(theta)+cos(theta)+1)/(cos(theta)sqrt(1-cos^2theta))#

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