Express (picture below) in terms of sin θ?

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Please show work and/or explain! :]

2 Answers
mizoo
Mar 15, 2018

#sqrt ((tan^2 θ) / (sec^2 θ + cot^2 θ sec^2 θ)) = sin^2 θ#

where #θ ≠ kpi + pi/2, kpi - pi/4#, k any integer

Explanation:

#sqrt ((tan^2 θ) / (sec^2 θ + cot^2 θ sec^2 θ))#

#sqrt ((sin^2 θ/cos ^2 θ) / (sec^2 θ( 1 + cot^2 θ))#

#sqrt ((sin^2 θ/cos ^2 θ) / (1/cos^2 θ( 1 + cot^2 θ))#

#sqrt ((sin^2 θ) / (( 1 + cot^2 θ))#

#=sqrt (sin^2 θ / ( ( 1 + (1/sin^2θ - 1 )#

# = sqrt (sin^2 θ / (1/sin^2 θ )) #

# = sqrt (sin^4 θ ) #

# = sin^2 θ # , where #cos ^2 θ and 1 + cot^2 θ≠0#

#cos ^2 θ = 0, θ = kpi + pi/2#, k any integer
#1 + cot^2 θ = 0, θ = kpi - pi/4#, k any integer

Note: How #cot^2 θ # became # 1/sin^2θ - 1# in the third line:

#sin^2 θ+ cos^2 θ = 1#
#sin^2 θ/sin^2 θ+ cos^2 θ/sin^2 θ = 1/sin^2 θ#
#1 + cot^2θ = 1/sin^2 θ#
#cot^2θ = 1/sin^2 θ - 1#

Jim G.
Mar 15, 2018

#sin^2theta#

Explanation:

#"let's begin by simplifying the contents of the radical"#

#"using the "color(blue)"trigonometric identities"#

#•color(white)(x)tantheta=sintheta/costheta" and "cottheta=costheta/sintheta#

#•color(white)(x)sectheta=1/costheta" and "sin^2theta+cos^2theta=1#

#rArr(tan^2theta)/(sec^2theta+cot^2thetasec^2theta)#

#=(sin^2theta/cos^2theta)/(1/cos^2theta+cancel(cos^2theta)/sin^2thetaxx1/cancel(cos^2theta)#

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

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

#=(sin^2theta/cos^2theta)/(1/(cos^2thetasin^2theta)#

#=sin^2theta/cancel(cos^2theta)xxcancel(cos^2theta)sin^2theta#

#=sin^4theta#

#rArrsqrt((tan^2theta)/(sec^2theta+cot^2thetasec^2theta)#

#=sqrt(sin^4theta)=sin^2theta#

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