Tricky Tricky little Identities

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The discussion revolves around solving a challenging trigonometric identity: 1/(1+sinx) + 1/(1-sinx) = 2sec^2x. The original poster expresses difficulty in proving the identity after a period of not practicing. A participant provides a solution, simplifying the left side by combining fractions and using the identity 1 - sin^2(x) = cos^2(x) to reach the right side. The original poster acknowledges a missed step in the simplification process but appreciates the assistance. The conversation highlights the importance of revisiting foundational concepts in trigonometry for problem-solving.
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My friend sent me some Trigonometric proving identity questions to practise and i am usually good at them but i haven't done them for a while so i have gotten a bit rusty plus these ones to me are very difficult so i would like some assistance.
Prove 1/1+sin + 1/1-sinx = 2secsquaredx.
ok so left side looks hardest so i started with that and did it first. It may look confusing writing the divisions.

1/1+sinx + 1/1-sinx
1/1+ cosx/cotx + 1/cosx
cotx+1/cosx + 1/cosx
cotx+1/cosx + 1
cscx/cosx +1
Ok so that is where i have gotten to and it doesn't seem like i can get it to equal the right side. I think i have screwed up a step and i would like any assistance if possible. Please help
 
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You are making it look harder than it is... Just a 3 step problem.

\frac{1}{1+sinx} + \frac{1}{1-sinx} = \frac{2}{cos^2(x)}
\frac{2}{(1+sinx)(1-sinx)} = ...
\frac{2}{1 - sin^2(x)} = \frac{2}{cos^2(x)} = 2sec^2(x)
 
Oh you skipped step where you make bases same but you it works, thankyou very much.
 
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