Question #0042b

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Answer 1 · 2017-08-24T22:22:43

The solution is: all #theta# except of # theta = pi/2+pik# with #k# an integer.

#sec^2 theta = 1/cos^2 theta# is not defined for #cos theta = 0#, so it is not defined for #theta = pi/4 +pik# with #k# an integer.

For all other #theta#, we have

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

So, every other #theta# is a solution.

Answer 2 · 2017-08-24T22:22:53

See explanation below.

We have: #(1 - sin^(2)(theta)) sec^(2)(theta)#

One of the Pythagorean identities is #cos^(2)(theta) + sin^(2)(theta) = 1#.

We can rearrange it to get:

#Rightarrow cos^(2)(theta) = 1 - sin^(2)(theta)#

Let's apply this rearranged identity to our proof:

#= cos^(2)(theta) cdot sec^(2)(theta)#

Let's apply the standard trigonometric identity #sec(theta) = frac(1)(cos(theta))#:

#= cos^(2)(theta) cdot frac(1)(cos^(2)(theta))#

#= frac(cos^(2)(theta))(cos^(2)(theta))#

#= 1 " " " # #"Q.E.D."#