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(1+sinx/cosx)+(cosx/1+sinx)=2/cosx Prove identity
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did you try cross-multiplying ? like, \(\Large \dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+cb}{bd}\)
if you tried that, then what would you get on the left side ?
(1+sinx)(1+sinx)+cosx.cosx/[cosx.(1+sinx)] [(1+sinx)^2+cos^2x]/cosx.(1+sinx) 1+sin^2x +2sinx+cos^2x /cosx(1+sinx) 1+1+2sinx/cosx(1+sinx) 2+2sinx/cosx(1+sinx) taking 2 as common and cancelling 1+sinx 2(1+sinx)/cosx(1+sinx) =2/cosx
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