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$cos^{-1}cos(-5)+$

$sin^{-1}sin6$-

$tan^{-1}tan12 =? $

$cos^{-1}(cos(-5))+sin^{-1}(sin(6))-tan^{-1}(tan(12)) $ is equal to: 

(The inverse trigonometric functions take the principal values)

(A) $3\pi - 11$
(B) $3\pi + 1 $
(C) $4\pi - 11$
(D) $4\pi - 9$

Solution

The given expression can be rewritten as,

$cos^{-1}cos5+sin^{-1}sin 6-tan^{-1} tan12 $

Considering the principal values the given expression can be further rewritten as,

$cos^{-1}cos(2\pi - 5)+sin^{-1}sin (-(2\pi- 6))-tan^{-1} tan(-(4\pi -12)) $

Or $(2\pi - 5)+ (-(2\pi- 6))-(-(4\pi-12)) = 4\pi -11 $

Answer: (C)

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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

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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