CONCEPT:

• For the function to be invertible for the given values of the dependent variable x, it should be Bijective for those values of x.

EXPLANATION:

• Referring to the graph of y = sin x
• For \(x ∈ \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\)

• For \(x ∈ \left[ - \frac{\pi}{2} , \frac{\pi}{2} \right]\), the function y = sin x is one-one as for each input value of x there will be only one value of y. In another word, if you will draw a horizontal line at any point in between \(x ∈ \left[ - \frac{\pi}{2} , \frac{\pi}{2} \right]\), there will be only one Intersection point so it is one-one
•  For the function to be onto,

Range = Co-domain of y = sin x = [-1, 1] for all \(x ∈ \left[ - \frac{\pi}{2} , \frac{\pi}{2} \right]\)

• So it is onto also.
• Finally for \(x ∈ \left[ - \frac{\pi}{2} , \frac{\pi}{2} \right]\) ⇒ y = sin x is Bijective ⇒ Invertible.
• So, the correct answer is option 3

CONCEPT:

for the function y = sin-1 x to exist,

⇒ |x| ≤ 1

CALCULATION:

For \(\sin^{-1} \left( 2 \sin \frac{\pi}{2} \right)\) to exist,

\(\Rightarrow -1 ≤ 2 \sin \frac{\pi}{2} ≤ 1\)

⇒ -1 ≤ 2 ≤ 1 ⇒ can't be true

⇒ The value of \(\sin^{-1} \left( 2 \sin \frac{\pi}{2} \right)\) will not exist.

So the correct answer is option 4.

Concept:

cos (-θ) = cos θ 

\(\cos\left({\pi\over2}\right) = 0\)

Calculation:

x = \(\rm \tan^{-1}\left[\cos\left(-{\pi\over2}\right)\right]\)

x = \(\rm \tan^{-1}\left[\cos\left({\pi\over2}\right)\right]\)

x = \(\rm \tan^{-1}\left[0\right]\)

∵ Principle value can be ∈ \(\left[-{\pi\over2},{\pi\over2}\right]\)

∴ x = 0

Concept:

sin x = y then x = sin^-1 y

\(\rm tan^{-1} \ x + cot^{-1} \ x = \frac {π}{2}\)

Calculation:

Given: \(\rm sin (tan^{-1}\ \frac {1}{10} \ + \ cot^{-1} \ x) = 1\)

⇒ \(\rm tan^{-1}\ \frac {1}{10} \ + \ cot^{-1} \ x = sin^{-1}\ (1)\) -----(∵ sin^-1 (1) = sin^-1 (sin (π/2)) = π/2)

⇒  \(\rm tan^{-1}\ \frac {1}{10} \ + \ cot^{-1} \ x = {\pi\over2}\)

Here, \(\rm tan^{-1} \ x + cot^{-1} \ x = \frac {π}{2}\)

Then x = \(\rm 1\over10\)

Additional Information

 \(\rm sin^{-1} \ x + cos^{-1} \ x = \frac {π}{2}\)

\(\rm cosec^{-1} \ x +sec^{-1} \ x = \frac {π}{2}\)

Concept:

• 1 + tan² x = sec² x.
• tan (tan^-1 x) = x.

Calculation:

Let's say that \(\rm \tan^{-1}\dfrac{y}{2}=\theta\), where \(\rm \theta \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\).

\(\rm \Rightarrow \tan \theta = \dfrac{y}{2}\)

\(\rm \Rightarrow \tan^2 \theta = \dfrac{y^2}{4}\)

\(\rm \Rightarrow 1+\tan^2 \theta = 1+\dfrac{y^2}{4}=\dfrac{y^2+4}{4}\)

\(\rm \Rightarrow \sec^2 \theta = \dfrac{y^2+4}{4}\)

\(\rm \Rightarrow \sec \theta = \dfrac{\sqrt{y^2+4}}{2}\)

\(\rm \Rightarrow \sec \left(\tan^{-1}\dfrac{y}{2}\right) = \dfrac{\sqrt{y^2+4}}{2}\).

Concept:

The domain of inverse sine function, sin x is \(x \in \left[ { - 1,1} \right]\)

Calculation:

The domain of the function \({\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)\) is calculated as follows:

\(1 - {x^2} \geq 0\)

⇒ \(x \in \left[ { - 1,1} \right]\)      ...(1)

Also, 

\(- 1 \le 2x\sqrt {1 - {x^2}} \le 1\)

\( - \frac{1}{2} \le x\sqrt {1 - {x^2}} \le \frac{1}{2}\)

Square it to get x,

\(0 \leq {x^2}\left( {1 - {x^2}} \right) \le \frac{1}{4}\)

Now, 

\( {x^2}\left( {1 - {x^2}} \right) \geq 0\) and \({x^2}\left( {1 - {x^2}} \right) \le \frac{1}{4}\)

Here we have to find the common values of x.

For, \( {x^2}\left( {1 - {x^2}} \right) \geq 0\)

Here, the values of x for which LHS will change its sign will be -1 and 1 so the values of x for the above inequality,

⇒ \(x \in \left[ { - 1,1} \right]\)      ....(2)

For,

\({x^2}\left( {1 - {x^2}} \right) \le \frac{1}{4}\)

\(t - {t^2} - \frac{1}{4} \le 0\)

\({\left( {t - \frac{1}{2}} \right)^2} \le 0\)

\(t \le \frac{1}{2}\)

\({x^2} \le \frac{1}{{\sqrt 2 }}\)

\(x \in \left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]\)        ....(3)

Take, all the common intervals from equations 1, 2, and 3,

We will get, 

\(⇒ x \in \left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]\)

Concept:

• The domain of a function is the set of values that we are allowed to plug into our function.
   The range is the set of all possible values that the function will give when we give in the domain as input.

Calculation:

• The above graph is of the function sec^-1x.
• Since secant function restricted to any of the intervals [-π, 0]-{-π/2}, [0, π ] - {π/2} etc. is bijective and its range is R - (-1, 1)
• Thus sec^-1 can be defined as a function whose domain is R - (-1, 1) and the range could be any of the intervals [-π, 0]-{-π/2}, [0, π ] - {π/2} etc., 
• Corresponding to each of these intervals, we get different branches of the function sec^-1, and the branch [0, π]-{\(\frac{π}{2}\)}, is called the principal value branch of the function sec^-1.
• We thus have sec^-1: R - (-1, 1) → [0,π] - {\(-\frac{π}{2}\)} 

∴ Domain(D) = R - (-1, 1) and range(R) =  [0,π] - {\(-\frac{π}{2}\)}

Hence, the correct answer is option 1).

Concept:

• Principal value: The value of an inverse trigonometric function that lies in its principal value branch is called the principal value of that inverse trigonometric function.
• The range of the principal value branch of tan^-1 is \(\big(\frac{-\pi}{2},\frac{\pi}{2}\big)\)
• The range of the principal value branch of sec-1 is \([0, \pi ]- {\frac{\pi}{2}}\)
• tan^-1(- x) = - tan^-1x

Calculations:

Let y = tan^-1(-√3)= - tan^-1(√3),

Then, tan y = -√3

We know that the range of the principal value branch of tan-1 is \(\big(\frac{-\pi}{2},\frac{\pi}{2}\big)\)

and \(tan (\frac{\pi}{3}) = \sqrt3\).

∴ the principal value of  tan-1(√3) is \(-\frac{\pi}{3}\) .......(1)

Next, 

Let x = sec-1(\(\frac{2}{\sqrt3}\)),

Then, sec x = \(\frac{2}{\sqrt3}\)

We know that the range of the principal value branch of sec-1 is \([0, \pi ]- {\frac{\pi}{2}}\)

and.

∴ the principal value of  sec-1(\(\frac{2}{\sqrt3}\)) is \(\frac{\pi}{3}\) .......(2)

Adding (1) and (2), we have

\(tan^{-1}(-\sqrt3) + 2 sec^{-1}(\frac{2}{\sqrt 3})= -\frac{\pi}{3}+2\times\frac{\pi}{6}\)

\(=0\)

∴ \(tan^{-1}(-\sqrt3) + 2 sec^{-1}(\frac{2}{\sqrt 3})= 0\)

Hence, the correct answer is option 2).

Calculation:

∵  sin ^{- 1}(1/2 x) = π/2 - cos ^{- 1}(1/2 x)

Given

cos - 1x + sin - 1(1/2 x) = π/6

⇒ cos ^{- 1}x + π/2 - cos ^{- 1}(1/2 x) = π/6

⇒ cos ^{- 1}x - cos ^{- 1}(1/2 x) = π/6 - π/2

⇒ cos ^{- 1}x = cos ^{- 1}(1/2 x) - π/3

⇒ cos ^{- 1}x = cos ^{- 1}(1/2 x) - cos ^{- 1}(1/2)

⇒ cos - 1x = cos ^{- 1}[x/2.1/2 + √(1 - x²/4).√(1 - 1/4)]

⇒ x = x/4 + √3/2 √(1 - x²/4)

⇒ 3/4 x = √3/4 √(4 - x²)

⇒ 3x² = 4 - x²

⇒ 4x² = 4

⇒ x = ±1

Concept:

\({\sin ^{ - 1}}(\sin {\rm{\theta }}) = {\rm{\theta }},{\rm{\;\;\;\;\;\theta }} \in \left( { - \frac{{\rm{\pi }}}{2},{\rm{\;}}\frac{{\rm{\pi }}}{2}} \right)\)

sin (2nπ + θ) = sin θ

sin (π – θ) = sin θ

Calculation:

We have to find the value of \({\sin ^{ - 1}}\sin \left( {\frac{{33{\rm{\pi }}}}{5}} \right)\)

\({\sin ^{ - 1}}\sin \left( {\frac{{33{\rm{\pi }}}}{5}} \right) = {\rm{\;}}{\sin ^{ - 1}}\sin \left( {6{\rm{\pi }} + \frac{{3{\rm{\pi }}}}{5}} \right) \)

\(= {\sin ^{ - 1}}\sin \left( {\frac{{3{\rm{\pi }}}}{5}} \right) \)                                  (∵ sin (2nπ + θ) = sin θ)

\(= {\rm{\;}}{\sin ^{ - 1}}\sin \left( {{\rm{\pi }} - \frac{{2{\rm{\pi }}}}{5}} \right) \)

\(= {\rm{\;}}{\sin ^{ - 1}}\sin \left( {\frac{{2{\rm{\pi }}}}{5}} \right) \)                                (∵ sin (π – θ) = sin θ)

Here \(\frac{{2{\rm{\pi }}}}{5}\) is lies between \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\),

\(\therefore {\rm{\;}}{\sin ^{ - 1}}\sin \left( {\frac{{2{\rm{\pi }}}}{5}} \right) \)\(= \frac{{2{\rm{\pi }}}}{5}\)