Trig
posted by McKenna Louise .
Prove the following:
1/(tanØ  secØ ) + 1/(tanØ + secØ) = 2tanØ
(1  sinØ)/(1 + sinØ) = sec^2Ø  2secØtanØ + tan^2Ø

That's a better job of typing it.
I did the first of these in your previous post..
the 2nd:
LS = (1sinØ)/(1+ sinØ)
= (1sinØ)/(1+ sinØ) * (1sinØ)/(1 sinØ)
= (1  2sinØ + sin^2 Ø)/(1  sin^2 Ø)
= (1  2sinØ + sin^2 Ø)/cos^2 Ø
= 1/cos^2 Ø  2sinØ/cos^2 Ø + sin^2 Ø/cos^2Ø
= sec^2 Ø  2(sinØ/cosØ)*(1/cosØ) + tan^2 Ø
= sec^2 Ø  2tanØ secØ + tan^2 Ø
= RS 
or, if you divide top and bottom by cosØ you have
(secØtanØ)/(secØ+tanØ)
now multiply top and bottom by (secØtanØ) and you have
(secØtanØ)^2 / (sec^2Øtan^2Ø)
= sec^2Ø  2secØtanØ + tan^2Ø 
Thanks to the both of you :)

hj
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