Question

tant 3-sei (-2) is equal to
.
$tan {?}^{ - 1} \sqrt{3} - sec {?}^{ - 1} ( - 2) \: is \: equal \: to$
​

Answer 1

AnswEr :

Given Expression,

$\sf tan {}^{ - 1}( \sqrt{3} ) - sec {}^{ - 1} ( - 2)$

We know that,

$\sf { sec}^{ - 1} ( - x) = \pi - {sec}^{ - 1} (x)$

Therefore,

$\implies \: \sf tan {}^{ - 1}( \sqrt{3}) + \pi \: - sec {}^{ - 1} (2)$

We know that,

• tan(π/3) = √3
• sec(π/3) = 2

Thus,

$\implies \: \sf tan {}^{ - 1} \big(tan( \dfrac{\pi}{3} ) \big) + \pi \: - sec {}^{ - 1} \big( sec( \dfrac{\pi}{3} ) \big) \\ \\ \implies \sf \pi + \dfrac{\pi}{3} - \dfrac{\pi}{3} \\ \\ \implies \sf \: \pi$

Henceforth,

$\boxed{ \boxed{\sf tan {}^{ - 1}( \sqrt{3} ) - sec {}^{ - 1} ( - 2) = \pi}}$

Answer 2

Step-by-step explanation:

AnswEr :

Given Expression,

\sf tan {}^{ - 1}( \sqrt{3} ) - sec {}^{ - 1} ( - 2)tan

−1

(

3

)−sec

−1

(−2)

We know that,

\sf { sec}^{ - 1} ( - x) = \pi - {sec}^{ - 1} (x)sec

−1

(−x)=π−sec

−1

(x)

Therefore,

\implies \: \sf tan {}^{ - 1}( \sqrt{3}) + \pi \: - sec {}^{ - 1} (2)⟹tan

−1

(

3

)+π−sec

−1

(2)

We know that,

tan(π/3) = √3

sec(π/3) = 2

Thus,

\begin{gathered}\implies \: \sf tan {}^{ - 1} \big(tan( \dfrac{\pi}{3} ) \big) + \pi \: - sec {}^{ - 1} \big( sec( \dfrac{\pi}{3} ) \big) \\ \\ \implies \sf \pi + \dfrac{\pi}{3} - \dfrac{\pi}{3} \\ \\ \implies \sf \: \pi\end{gathered}

⟹tan

−1

(tan(

3

π

))+π−sec

−1

(sec(

3

π

))

⟹π+

3

π

−

3

π

⟹π

Henceforth,

\boxed{ \boxed{\sf tan {}^{ - 1}( \sqrt{3} ) - sec {}^{ - 1} ( - 2) = \pi}}

tan

−1

(

3

)−sec

−1

(−2)=π