Question #f4b07

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Answer 1 · 2017-03-21T00:16:09

I tried this:

Have a look:
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Answer 2 · 2017-03-21T00:38:02

Please see the proofs below.

Verify:

#cos(u)/(1-sin(u))=(1+sin(u))/cos(u)#

Multiply the left side by 1 in the form of #(1 + sin(u))/(1 + sin(u))#:

#(1 + sin(u))/(1 + sin(u))cos(u)/(1-sin(u))=(1+sin(u))/cos(u)#

The denominator becomes the difference of two squares:

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

Substitute #cos^2(u)# for #(1-sin^2(u))#:

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

The cosine in the numerator cancels one of the cosines in the denominator:

#(1 + sin(u))/cos(u)=(1+sin(u))/cos(u)#

Verified.

Verify:

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

Multiply the left side by 1 in the form #(sec(x)-1)/(sec(x)-1)#

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

The numerator becomes the difference of two squares:

#(sec(x)-1)tan^2(x)/(sec^2(x)-1) = sec(x)-1#

Substitute #tan^2(x)# for #sec^2(x)-1#

#(sec(x)-1)tan^2(x)/(sec^2(x)-1) = sec(x)-1#

The fraction becomes 1:

#sec(x)-1 = sec(x)-1#

Verified.