If tan^{-1}(x+h)=tan^{-1}x+(hsiny)(siny)-(hsiny)^2.\frac{sin2y}{2}+(hsiny)^3.\frac{sin3y}{3}+..........\infty
if x\epsilon (0,1),y\epsilon (\frac{\pi }{4},\frac{\pi }{2})
then y = ?
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1 Answers
66The given equation is equivalent to
\dfrac{\tan^{-1}(x+h)-\tan^{-1}x}{h}=\sin^2y+h\times \mathrm{something\ finite}
Taking the limit as h→0, we get
\dfrac{\mathrm d}{\mathrm dx}\tan^{-1}x=\sin^2y
That is \sin^2y=\dfrac{1}{1+x^2}
Hence,
\sin y=\dfrac{1}{\sqrt{1+x^2}}
P.S: By the way, use \sin y to get \sin y and NOT siny to get sin y