Q.10  limit as x rightwards arrow pi divided by 2 of space left square bracket space 1 minus tan left parenthesis x divided by 2 right parenthesis right square bracket left square bracket 1 minus space sin x space right square bracket space space divided by space space left square bracket space space 1 plus tan left parenthesis x divided by 2 right parenthesis space left square bracket pi minus 2 x right square bracket 3  is    (1)  1/8  (2)  0  (3) 1/32 (4) infinity 

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limxπ21-tanx21-sinx1+tanx2π-2x3Then the answer isFirst we have to simplify 1-tanx21+tanx2Multiplying both numerator and denominator with cosx2 we getcosx2-sinx2cosx2+sinx2now rationalizing the by multiplying with cosx2-sinx2we get1-tanx21+tanx2=cosx2-sinx2cosx2+sinx2×cosx2-sinx2cosx2-sinx2=cosx2-sinx22cos2x2-sin2x2=cos2x2+sin2x2-2sinx2cosx2cosx=1-sinxcosxThus the question simplifies to1-sinx2cosxπ-2x3Let π-2x=txπ2,t0also x=π2-t2cosx=cosπ2-t2=sint2=2sint4cost4 Also 1-sinx=1-sinπ2-t2=1-cost2=2sin2 t41-sinx 2=4sin4t4Tha above equation transforms tolimt04sin4t42sint4cost4t32limt0sin3t4cost4t32limt0sin3t4t3                as t0 cost4=cos0=1    2143264=132  

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