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Find d/dx tan -1 √1+sinx/1-sins; 0

Find d/dx tan-1 √1+sinx/1-sins; 0

Grade:12

1 Answers

Y RAJYALAKSHMI
45 Points
7 years ago
tan-1 √1+sinx/1-sinsx
= tan-1 √(cos2x/2 + sin2x/2 + 2*sinx/2*cosx/2)/(cos2x/2 + sin2x/2 – 2*sinx/2*cosx/2)
=tan-1 √(cosx/2 + sinx/2)2/(cosx/2  –  sinx/2)2
tan-1 (cosx/2 + sinx/2)/(cosx/2  –  sinx/2)
= tan-1 (1 + tanx/2)/(1  –  tanx/2)     (by dividing the numerator & denominator by cos x/2)
tan-1 (tan(pi/4) + x/2)
= (pi/4) + x/2
Hence d/dx tan-1 √1+sinx/1-sinx = d/dx [(pi/4) + x/2] = 1/2
 
Ans: 1/2
 

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