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In a triangle ABC, let $AB=\sqrt {23}$, $BC=3$ and ....

In a triangle ABC, let $AB=\sqrt {23}$, $BC=3$ and $CA=4$. Then the value of

$$\frac {cot A + cot C}{cot B}$$

is _______ .

Solution

$\frac {cot A + cot C}{cot B}=\frac {sin (A+C).sin B}{sin A sin C.cos B}$

$= \frac {sin^2 B}{sin A. sin C. cos B}$

$= \frac {b^2}{a.c. \frac {c^2+a^2-b^2}{2ac}}$

$= \frac {2b^2}{c^2+a^2-b^2}$

Putting $AB = c = \sqrt {23},BC=a=3,CA=b=4$ we have, $ \frac {2.4^2}{23+3^2-4^2}$

$= \frac {2\times 16}{23+9-16}=\frac {2\times 16}{16}=2$

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The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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