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For Maths Lovers...

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prove the identity....

tanx + cotx + secx + cosecx (is identical to) .. (sinx+1)(cosx+1)secxcosecx-1
 
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solved that identity....will post soln later...u try this...
 

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umarashraf said:
prove the identity....

tanx + cotx + secx + cosecx (is identical to) .. (sinx+1)(cosx+1)secxcosecx-1

Left Hand Side=tanx + cotx + secx + cosec x
= sinx/cosx + cosx/sinx + 1/cosx + 1/sinx
=(sinx + 1)/cosx + (cosx + 1)/sinx
= (sinxsinx + sinx + cosxcosx + cosx)/cosxsinx
= (sinx + cosx + 1)/cosxsinx
= (cosxsinx + sinx + cosx +1-cosxsinx)/cosxsinx
= (cosxsinx + sinx + cosx + 1)/cosxsinx - (cosxsinx/cosxsinx)
= {sinx(cosx+1)+(cosx+1)}/cosxsinx - 1
= (sinx + 1)(cosx + 1)?cosxsinx - 1
= (sinx + 1)(cosx + 1)secxcosecx - 1
= R.H.S
Hence proved

I feel there is a shorter route through this but I can't figure it out. I think it may be solved Without simplifying all terms into sin and cos but I can't! Anyone?
 
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if u substitute the value of A too calculate C u ll find a negative anser.....which is impossible for a length so its B !!!
:lol: :lol: :lol:
 
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(sinx+1) (cosx+1)secx.cosecx-1
= secx.cosecx(sinx(cosx+1)+1(cosx+1))-1
= secx.cosecx(sinx.cosx+sinx+cosx+1)-1
= secx.cosecx.sinx.cosx+secx.cosecx.sinx+secx.cosecx.cosx+secx.cosecx-1
= 1+(1/cosx)+(1/sinx)+(1/sinx.cosx)-1
= (sinx+cosx+1)/sinx.cosx
= (sinx+cosx+sin^2 x + cos^2 x)/sinx.cosx
= (sinx/sinx.cosx)+(cosx/sinx.cosx)+(sin^2 x/sinx.cosx) + (cos^2 x/sinx.cosx)
= secx + cosecx + tanx + cotx
 
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:ROFLMAO: :crazy: am going nuts...i dreamt abt this thread last nite...just remembered.
For the proof part,
tanx + cotx+secx+cosecx
=sinx/cosx + cosx/sinx + 1/cosx + 1/sinx
= (sin^2 x + cos^2 x + sinx + cos x) / sinxcosx
= (1+sinx + cosx)/ sinxcosx
= (1+sinx +cosx+ sinxcosx -sinxcosx)/sinxcosx
= (sinx + 1)(cosx + 1)/sinxcosx - (sinxcosx/sinxcox)
=(sinx+1) (cosx+1)secxcosecx-1
Done.
Done!

@Hassam, lol. common sense betrayed me.
@Yukified, different approach! :)
 
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