Question #ae7ae

2 Answers
P dilip_k
Jan 27, 2018

#LHS=(tan theta+sec theta - 1)/ (tan theta -sec theta + 1)#

#=(tan theta+sec theta - (sec^2theta-tan^2theta))/ (tan theta -sec theta + 1)#

#=(tan theta+sec theta + (tan^2theta-sec^2theta))/ (tan theta -sec theta + 1)#

#=((tan theta+sec theta )(1+ tantheta-sectheta))/ (tan theta -sec theta + 1)#

#=tan theta+sec theta#

#=sin theta/costheta+1/costheta#

#=(sin theta+1)/costheta=RHS#

Sean
Jan 27, 2018

Please see below.

Explanation:

.

#(tantheta+sectheta-1)/(tantheta-sectheta+1)=(1+sintheta)/costheta#

#(sintheta/costheta+1/costheta-1)/(sintheta/costheta-1/costheta+1)=((sintheta+1-costheta)/costheta)/((sintheta-1+costheta)/costheta)=((sintheta+1)-costheta)/(sintheta+costheta-1)=#

#([(sintheta+1)-costheta][(sintheta+1)+costheta])/([(sintheta+costheta)-1][(sintheta+costheta)+1])=#

#([(sintheta+1)^2-cos^2theta])/([(sintheta+costheta)^2-1])=#

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

#(2sintheta+2sin^2theta)/(2sinthetacostheta)=(2sintheta(1+sintheta))/(2sinthetacostheta)=(cancelcolor(red)(2sintheta)(1+sintheta))/(cancelcolor(red)(2sintheta)costheta)=#

#(1+sintheta)/costheta#

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