A) \[\frac{2}{1+{{t}^{2}}}\]
B) \[\frac{1}{1+{{t}^{2}}}\]
C) 1
D) 2
Correct Answer: C
Solution :
\[\tan y=\frac{2t}{1-{{t}^{2}}}\] .....(i) and \[\sin x=\frac{2t}{1+{{t}^{2}}}\] .....(ii) From (i), differentiating w.r.t. t of y, we get, \[\frac{dy}{dx}=\frac{1}{2}\frac{\cos (\sqrt{\sin x+\cos x})}{\sqrt{\sin x+\cos x}}(\cos x-\sin x)\] and \[\frac{dy}{dt}=\frac{2(1+{{t}^{2}})}{{{(1-{{t}^{2}})}^{2}}}.\frac{1}{(1+{{\tan }^{2}}y)}\] or \[\frac{dy}{dt}=\frac{2(1+{{t}^{2}})}{{{(1-{{t}^{2}})}^{2}}}.\frac{1}{\left[ 1+{{\left( \frac{2t}{1-{{t}^{2}}} \right)}^{2}} \right]}=\frac{2}{1+{{t}^{2}}}\] .....(iii) and from (ii), differentiating w.r.t. t of x, we get \[\cos x\frac{dx}{dt}=\frac{2(1-{{t}^{2}})}{{{(1+{{t}^{2}})}^{2}}}\] or \[\frac{dx}{dt}=\frac{2(1-{{t}^{2}})}{{{(1+{{t}^{2}})}^{2}}}\frac{1}{\sqrt{1-\frac{{{(2t)}^{2}}}{{{(1+{{t}^{2}})}^{2}}}}}=\frac{2}{1+{{t}^{2}}}\] .....(iv) Hence \[\frac{dy}{dx}=1\].
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