The number of integral values of $k$ for which the equation $s i n^{- 1} x + t a n^{- 1} x = 2 k + 1$ has a solution is

1

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2

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3

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4

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Solution

The correct option is B $2$
$- 1 \leq x \leq 1$
Here $s i n^{- 1} (x)$ c values can lie between -1 to 1 so $t a n^{- 1} (x)$ domain is also restricted to -1 to 1.
Hence the range of the given expression is $\frac{- 3 π}{4} \leq 2 k + 1 \leq \frac{3 π}{4}$
Therefore possible values of k are -1,0.

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