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Question: If \[\left[ {{{\sin }^{ - 1}}\left( {{{\cos }^{ - 1}}\left( {{{\sin }^{ - 1}}\left( {{{\tan }^{ - 1}...

If [sin⁡−1(cos⁡−1(sin⁡−1(tan⁡−1x)))]=1,\left[ {{{\sin }^{ - 1}}\left( {{{\cos }^{ - 1}}\left( {{{\sin }^{ - 1}}\left( {{{\tan }^{ - 1}}x} \right)} \right)} \right)} \right] = 1, where [.]\left[ . \right] denotes the greatest integer function then the values of x are
A. [tan⁡sin⁡cos⁡1,tan⁡sin⁡cos⁡sin⁡1]\left[ {\tan \sin \cos 1,\tan \sin \cos \sin 1} \right]
B. (tan⁡sin⁡cos⁡1,tan⁡sin⁡cos⁡sin⁡1)\left( {\tan \sin \cos 1,\tan \sin \cos \sin 1} \right)
C. [−1,1]\left[ { - 1,1} \right]
D. [sin⁡cos⁡tan⁡1,sin⁡cos⁡sin⁡tan⁡1]\left[ {\sin \cos \tan 1,\sin \cos \sin \tan 1} \right]

Explanation

Solution

Hint- Here in this problem the concept of inverse trigonometry and greatest function property has been used which will be explained first then accordingly we will come to the answer of the question. Greatest Integer function is defined as when the input is provided to the function the output is the integer which is less than or equal to the input provided. It is denoted by []\left[ { } \right]

Complete step-by-step answer:
For example
[3.5]=3\left[ {3.5} \right] = 3
The given function is
[sin⁡−1(cos⁡−1(sin⁡−1(tan⁡−1x)))]=1\left[ {{{\sin }^{ - 1}}\left( {{{\cos }^{ - 1}}\left( {{{\sin }^{ - 1}}\left( {{{\tan }^{ - 1}}x} \right)} \right)} \right)} \right] = 1
Now in order to solve, we will let define some variables
Let θ=cos⁡−1(sin⁡−1(tan⁡−1x))\theta = {\cos ^{ - 1}}\left( {{{\sin }^{ - 1}}\left( {{{\tan }^{ - 1}}x} \right)} \right)
Substitute this value in the above equation
[sin⁡−1θ]=1\left[ {{{\sin }^{ - 1}}\theta } \right] = 1
Also we know that from the definition of sin
1⩽θ⩽π21 \leqslant \theta \leqslant \dfrac{\pi }{2}
Therefor the above equation becomes
sin⁡1⩽θ⩽sin⁡π2 sin⁡1⩽θ⩽1  \sin 1 \leqslant \theta \leqslant \sin \dfrac{\pi }{2} \\\ \sin 1 \leqslant \theta \leqslant 1 \\\
Now, θ=cos⁡−1(sin⁡−1(tan⁡−1x))\theta = {\cos ^{ - 1}}\left( {{{\sin }^{ - 1}}\left( {{{\tan }^{ - 1}}x} \right)} \right)
Let ϕ=sin⁡−1(tan⁡−1x)\phi = {\sin ^{ - 1}}\left( {{{\tan }^{ - 1}}x} \right)
Therefore
θ=cos⁡−1(sin⁡−1(tan⁡−1x)) θ=cos⁡−1ϕ  \theta = {\cos ^{ - 1}}\left( {{{\sin }^{ - 1}}\left( {{{\tan }^{ - 1}}x} \right)} \right) \\\ \theta = {\cos ^{ - 1}}\phi \\\
Substitute the value of θ\theta in the equation sin⁡1⩽θ⩽1\sin 1 \leqslant \theta \leqslant 1
sin⁡1⩽cos⁡−1ϕ⩽1 cos⁡sin⁡1⩽ϕ⩽cos⁡1  \sin 1 \leqslant {\cos ^{ - 1}}\phi \leqslant 1 \\\ \cos \sin 1 \leqslant \phi \leqslant \cos 1 \\\
Again let β=tan⁡−1x\beta = {\tan ^{ - 1}}x
Therefore ϕ=sin⁡−1(tan⁡−1x)\phi = {\sin ^{ - 1}}\left( {{{\tan }^{ - 1}}x} \right)
ϕ=sin⁡−1β\phi = {\sin ^{ - 1}}\beta
Substitute the value of ϕ\phi in equation cos⁡sin⁡1⩽ϕ⩽cos⁡1\cos \sin 1 \leqslant \phi \leqslant \cos 1
cos⁡sin⁡1⩽sin⁡−1β⩽cos⁡1 sin⁡cos⁡sin⁡1⩽β⩽sin⁡cos⁡1  \cos \sin 1 \leqslant {\sin ^{ - 1}}\beta \leqslant \cos 1 \\\ \sin \cos \sin 1 \leqslant \beta \leqslant \sin \cos 1 \\\
As we let β=tan⁡−1x\beta = {\tan ^{ - 1}}x substitute it in the above equation we get
sin⁡cos⁡sin⁡1⩽tan⁡−1x⩽sin⁡cos⁡1 tan⁡sin⁡cos⁡sin⁡1⩽x⩽tan⁡sin⁡cos⁡1  \sin \cos \sin 1 \leqslant {\tan ^{ - 1}}x \leqslant \sin \cos 1 \\\ \tan \sin \cos \sin 1 \leqslant x \leqslant \tan \sin \cos 1 \\\
Therefore x lies between tan⁡sin⁡cos⁡sin⁡1⩽x⩽tan⁡sin⁡cos⁡1\tan \sin \cos \sin 1 \leqslant x \leqslant \tan \sin \cos 1
Hence the correct option is A.

Additional information-
The domain of function is all the values that are allowed as input in the function and range of the function is all the values that a function gives as an output.

Note- In order to solve these types of questions, you need to have a good concept of domain and range of a function. In the above question we define the range of arcsin and then proceed to solve the problem step by step. Learn the domain and range of all trigonometric functions and the concept of solving inequality problems.