Question

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Answer 1

y =tan−1 (1−sinx√+1+sin x/√1−sinx√−1+sin x√)1−sinx√+1+sin x/√1−sinx√−1+sin x√=sin2x/2+cos2x/2−2sinx/2cosx/2√+sin/2x/2+cos2x/2+2sinx/2cosx/2√sin2x/2+cos2x/2−2sinx/2cosx/2√−sin2x/2+cos2x/2+2sinx/2cosx/2√=(sinx/2−cosx/2)2√+(sinx/2+cosx/2)2√(sinx/2−cosx/2)2√−(sinx/2+cosx/2)2√=sinx/2−cosx/2+sinx2+cosx/2sinx/2−cosx/2−sinx/2−cosx/2=2sinx/2−2cosx/2= −tanx/2Now putting all these value we get, y =tan−1 (1−sinx√+1+sin x/√1−sinx√−1+sin x√)=tan−1(−tanx2)

or y = −tan−1(tanx2)=−x/2So dy/dx=−dd/xx2=−1/2