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Prove that- tan A/1-cot A + cot A/1-tan A=1+tan A+cot A
Asked by Aditya Chakravarty | 23 May, 2013, 12:44: PM
answered-by-expert Expert Answer
 
Answer : To prove :  tan A/1-cot A + cot A/1-tan A=1+tan A+cot A  
 
taking L.H.S 
= [ tan(A) / (1 - cot(A)) ] + [ cot(A) / (1 - tan(A) ) ] 

= [ sin(A)/cos(A) / (1 - cos(A)/sin(A)) ] + [ cos(A)/sin(A) / (1 - sin(A)/cos(A) ) ]
              {using tan x = sinx /cos x and cotx = cosx/sinx }

= [ sin(A)/cos(A) / (sin(A)/sin(A) - cos(A)/sin(A)) ] + [ cos(A)/sin(A) / (cos(A)/cos(A) - sin(A)/cos(A) ) ] 

= [ sin(A)/cos(A) / (sin(A) - cos(A)) / sin(A) ] + [ ( cos(A)/sin(A) / ( cos(A) -sin(A) ) / cos(A) ) ] 

= [ sin(A)sin(A) / cos(A)(sin(A) - cos(A)) ] + [ ( cos(A)cos(A) / sin(A)( cos(A) -sin(A) ) ]

=[ sin2(A) / cos(A)(sin(A) - cos(A)) ] + [ cos2(A) / -sin(A)( sin(A) - cos(A) ) ) ]

=[ sin2(A) / cos(A)(sin(A) - cos(A)) ] - [ cos2(A) / sin(A)( sin(A) - cos(A) ) ) ]               ===> LCD 

= [ sin2(A) sin(A) / cos(A)sin(A)(sin(A) - cos(A)) ] - [ cos2(A) cos(A) / sin(A)cos(A)( sin(A) - cos(A) ) ) ]

= [ sin3(A) / cos(A)sin(A)(sin(A) - cos(A)) ] - [ cos3(A) / sin(A)cos(A)( sin(A) - cos(A) ) ) ]

=[ sin3(A) - cos3(A) ] / [ sin(A)cos(A)( sin(A) - cos(A) ) ) ]

=[ (sin(A) - cos(A)) ( sin2(A) + sin(A) cos(A) + cos2(A) ) ] / [ sin(A)cos(A)( sin(A) - cos(A) ) ) ]

= [ ( sin2(A) + cos2(A) + sin(A) cos(A) ] / [ sin(A)cos(A) ]

= [ sin2(A) / sin(A)cos(A) ] + [ cos2(A) / sin(A)cos(A) ] + [ sin(A) cos(A) / sin(A) cos(A) ]

= [ sin(A) / cos(A) ] + [ cos(A) / sin(A) ] + 1

= tan(A) + cot(A) + 1
 {we know sin x/ cos x = tan x and cos x/sin x = cot x }
= R. H . S 
Hence proved
Answered by | 23 May, 2013, 08:06: PM

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