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PleaPl help me with the question given in the image attachment.

PleaPl help me with the question given in the image attachment. 

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Grade:12th pass

2 Answers

Arun
25750 Points
5 years ago
Lim x->a (cos x - cos a / cot x - cot a) 
= Lim x->a (cos x - cos a) / [(cos x sin a - cos a sin x)/(sin x sin a)] 
= Lim x->a (-2sin((x-a)/2)sin((x+a)/2)) / [(sin(a-x))/(sin x sin a)] 
= sin^3 a 
--------- 
Ideas: cos x - cos a = cos[(x+a)/2 + (x-a)/2] - cos[(x+a)/2 - (x-a)/2] = -2sin((x-a)/2)sin((x+a)/2)
Samyak Jain
333 Points
5 years ago
 lim   (cosx – cosa) / (cotx – cota)  It is of 0 by 0 form. Simplifying it we can calculate the limit.
x\rightarrowa
We know cosx – cosa = 2 sin(x+a)/2 sin(a – x)/2  and
cotx – cota = cosx/sinx – cosa/sina = (sina cosx – cosa sinx) / sinx sina = sin(a – x) / sinx sina
\therefore lim (cosx – cosa) / (cotx – cota) = lim [2 sin(x+a)/2 sin(a – x)/2] / [sin(a – x) / sinx sina]
 x\rightarrowa                                            x\rightarrowa
 = lim  [2 sin(x+a)/2 sin(a – x)/2 . sinx sina ] / [2 sin (a – x)/2 cos(a – x)/2 ]
  x\rightarrowa
= lim  sin(x+a)/2 . sinx sina / cos(a – x)/2  =  sina lim sin(x+a)/2 . sinx / lim cos(a – x)/2
 x\rightarrowa                                                                 x\rightarrowa                         x\rightarrowa
= sina sin(a+a)/2 . sina / cos(a – a)/2  =  sina sina .sina / cos0
= sin3a

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