Tan -1-3- -1-2 is equal toa -b -3c -3

Let tan^ 1√(3)=x⇒ tan x=√(3)⇒ tan x=tanπ/3 ⇒ x=π/3ϵ ( π/2,π/2 ) principal interval Let sec^ 1( 2)=y⇒ sec y= 2 ⇒ sec y= secπ/3⇒ sec y=sec ( π π/3 ) [∵ se ...

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Byju's Answer

Standard XII

Physics

Addition and Subtraction in Unit Vector Notation

tan -1√3- -1-...

Question

$t a n^{- 1} \sqrt{3} - s e c^{- 1} (- 2)$ is equal to
a) $π$
b) $- \frac{π}{3}$
c) $\frac{π}{3}$
d) $\frac{2 π}{3}$

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Solution

Let $t a n^{- 1} \sqrt{3} = x \Rightarrow t a n x = \sqrt{3} \Rightarrow t a n x = t a n \frac{π}{3}$
$\Rightarrow x = \frac{π}{3} ϵ (- \frac{π}{2}, \frac{π}{2})$ principal interval
Let $s e c^{- 1} (- 2) = y \Rightarrow s e c y = - 2$
$\Rightarrow s e c y = - s e c \frac{π}{3} \Rightarrow s e c y = s e c (π - \frac{π}{3}) [∵ s e c (π - θ) = - s e c θ]$
$\Rightarrow s e c y = s e c (\frac{2 π}{3}) \Rightarrow y = \frac{2 π}{3} = [0, π] - (\frac{π}{2})$ (principal interval)
$∴ t a n^{- 1} \sqrt{3} - s e c^{- 1} (- 2) = x - y = \frac{π}{3} - \frac{2 π}{3} = - \frac{π}{3}$

Hence, the correct option is (b)

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tan−1√3−sec−1(−2) is equal to a) π b) −π3 c) π3 d) 2π3

$t a n^{- 1} \sqrt{3} - s e c^{- 1} (- 2)$ is equal to
a) $π$
b) $- \frac{π}{3}$
c) $\frac{π}{3}$
d) $\frac{2 π}{3}$