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If $\sqrt{a+b}\tan\large\frac{\theta}{2}$$=\sqrt{a-b}\tan\large\frac{ \phi}{2}$ then $(b-a\sec \theta)(b+a\cos\phi)$ has maximum value $\lambda$ then $\lambda$ is

$(a)\;0\qquad(b)\;1\qquad(c)\;2\qquad(d)\;None\;of\;these$

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1 Answer

Given that
$\sqrt{a+b}\tan\large\frac{\theta}{2}$$=\sqrt{a-b}\sin\large\frac{\phi}{2}$
$\Rightarrow a\geq b$
$\cos\theta=\large\frac{1-\tan^2\Large\frac{\theta}{2}}{1+\tan^2\Large\frac{\theta}{2}}$
$\quad\quad\;=\large\frac{a\cos\phi+b}{a+b\cos\phi}$
$\Rightarrow (b-a\sec\theta)(b+a\cos\phi)=b^2-a^2$
$\Rightarrow (b-a\sec\theta)(b+a\cos\phi)\leq 0$
Maximum value is 0
$\therefore\lambda=0$
Hence (a) is the correct answer.
answered Oct 15, 2013 by sreemathi.v
 

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