EXAMPLE 3 140

JEE/Engineering Exams

Maths

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$$\begin{gathered} \frac{y+1}{1-y}=\sqrt{\frac{1+\sin \Theta }{1-{\sin }^2}} \\ =\sqrt{\frac{{\sin }^2\Theta /2+{\cos }^2\Theta /2+2\sin \Theta /2\cos \Theta /2}{{\sin }^2\Theta /2+{\cos }^2\Theta /2-2\sin \Theta /2\cos \Theta /2}} \\ =\sqrt{\frac{{(\cos \Theta /2+\sin \Theta /2)}^2}{{\left(\cos {\displaystyle \frac{\Theta }{2}}-\sin {\displaystyle \frac{\Theta }{2}}\right)}^2}} \\ =\frac{\cos \Theta /2+\sin \Theta /2}{\cos \Theta /2-\sin \Theta /2} \\ =\frac{1+\tan \Theta /2}{1-\tan \Theta /2} \\ Use compendo \& dividends rule \\ =\frac{y+1+1-y}{y+1-1+y}=\frac{1+\tan \Theta /2+1-\tan \Theta /2}{1+\tan \Theta /2-1+\tan \Theta /2} \\ \frac{2}{2y}=\frac{2}{2\tan \Theta /2} \\ \Rightarrow y=\tan \frac{\Theta }{2} \end{gathered}$$

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