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How do you verify the identity #sqrt((sinthetatantheta)/sectheta)=abs(sintheta)#?

Answer 1

Show that LHS = RHS for all #theta#

LHS simplifies to #sqrt ((sin(theta).sin(theta)/cos(theta))/(1/cos(theta))#using the definitions of tan and sec. This simplifies to #sqrt(sin(theta)^2# and hence to #sin (theta)#, where #sin(theta)# ≥ 0 and RHS = #sin(theta)#, where #sin(theta)# ≥ 0 Thus LHS = RHS for all values ot #theta#.

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Answer 2

Anna Andrews

Square LHS and simplify - as below.

#LHS = sqrt((sin theta tan theta)/sec theta)#

Consider the square of the LHS:

# LHS^2 = (sin theta tan theta)/sec theta#

#= sin thetaxxsin theta/cos theta xx cos theta#

#= sin thetaxxsin theta/cancel cos theta xx cancel cos theta# (#cos theta !=0 -> θ != pi/2 +npi # for all #n in ZZ#)

#= sin^2 theta#

#:. LHS^2 = sin^2 theta#

#LHS = +- sin theta = abs sin theta#

#LHS = RHS#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.