CBSE Class 12 Mathematics NCERT Solutions: Inverse Trigonometric Functions

Let the principal value of $\backslash sin^\{-1\}\backslash left(-\backslash frac\{1\}\{2\}\backslash right)$ be $y$. Then, by definition, $\backslash sin\; y\; =\; -\backslash frac\{1\}\{2\}$.

We know that $\backslash sin\backslash left(\backslash frac\{\backslash pi\}\{6\}\backslash right)\; =\; \backslash frac\{1\}\{2\}$. Therefore, $\backslash sin\; y\; =\; -\backslash sin\backslash left(\backslash frac\{\backslash pi\}\{6\}\backslash right)$.

Using the property $-\backslash sin(\backslash theta)\; =\; \backslash sin(-\backslash theta)$, we can write $\backslash sin\; y\; =\; \backslash sin\backslash left(-\backslash frac\{\backslash pi\}\{6\}\backslash right)$.

The range of the principal value branch for $\backslash sin^\{-1\}(x)$ is $\backslash left[-\backslash frac\{\backslash pi\}\{2\},\; \backslash frac\{\backslash pi\}\{2\}\backslash right]$. Since $-\backslash frac\{\backslash pi\}\{6\}$ lies within this range, it is the principal value.

Thus, the principal value of $\backslash sin^\{-1\}\backslash left(-\backslash frac\{1\}\{2\}\backslash right)$ is $-\backslash frac\{\backslash pi\}\{6\}$.

Let the principal value of $\backslash cos^\{-1\}\backslash left(\backslash frac\{\backslash sqrt\{3\}\}\{2\}\backslash right)$ be $y$. Then, by definition, $\backslash cos\; y\; =\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}$.

We know that $\backslash cos\backslash left(\backslash frac\{\backslash pi\}\{6\}\backslash right)\; =\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}$.

The range of the principal value branch for $\backslash cos^\{-1\}(x)$ is $\backslash left[0,\; \backslash pi\backslash right]$. Since $\backslash frac\{\backslash pi\}\{6\}$ lies within this range, it is the principal value.

Thus, the principal value of $\backslash cos^\{-1\}\backslash left(\backslash frac\{\backslash sqrt\{3\}\}\{2\}\backslash right)$ is $\backslash frac\{\backslash pi\}\{6\}$.

Let the principal value of $\backslash csc^\{-1\}(2)$ be $y$. Then, by definition, $\backslash csc\; y\; =\; 2$.

We know that $\backslash csc\backslash left(\backslash frac\{\backslash pi\}\{6\}\backslash right)\; =\; \backslash frac\{1\}\{\backslash sin\backslash left(\backslash frac\{\backslash pi\}\{6\}\backslash right)\}\; =\; \backslash frac\{1\}\{\backslash frac\{1\}\{2\}\}\; =\; 2$.

The range of the principal value branch for $\backslash csc^\{-1\}(x)$ is $\backslash left[-\backslash frac\{\backslash pi\}\{2\},\; \backslash frac\{\backslash pi\}\{2\}\backslash right]\; -\; \backslash \{0\backslash \}$. Since $\backslash frac\{\backslash pi\}\{6\}$ lies within this range, it is the principal value.

Thus, the principal value of $\backslash csc^\{-1\}(2)$ is $\backslash frac\{\backslash pi\}\{6\}$.

Let the principal value of $\backslash tan^\{-1\}(-\backslash sqrt\{3\})$ be $y$. Then, by definition, $\backslash tan\; y\; =\; -\backslash sqrt\{3\}$.

We know that $\backslash tan\backslash left(\backslash frac\{\backslash pi\}\{3\}\backslash right)\; =\; \backslash sqrt\{3\}$. Therefore, $\backslash tan\; y\; =\; -\backslash tan\backslash left(\backslash frac\{\backslash pi\}\{3\}\backslash right)$.

Using the property $-\backslash tan(\backslash theta)\; =\; \backslash tan(-\backslash theta)$, we can write $\backslash tan\; y\; =\; \backslash tan\backslash left(-\backslash frac\{\backslash pi\}\{3\}\backslash right)$.

The range of the principal value branch for $\backslash tan^\{-1\}(x)$ is $\backslash left(-\backslash frac\{\backslash pi\}\{2\},\; \backslash frac\{\backslash pi\}\{2\}\backslash right)$. Since $-\backslash frac\{\backslash pi\}\{3\}$ lies within this range, it is the principal value.

Thus, the principal value of $\backslash tan^\{-1\}(-\backslash sqrt\{3\})$ is $-\backslash frac\{\backslash pi\}\{3\}$.

Let the principal value of $\backslash cos^\{-1\}\backslash left(-\backslash frac\{1\}\{2\}\backslash right)$ be $y$. Then, by definition, $\backslash cos\; y\; =\; -\backslash frac\{1\}\{2\}$.

We know that $\backslash cos\backslash left(\backslash frac\{\backslash pi\}\{3\}\backslash right)\; =\; \backslash frac\{1\}\{2\}$. Therefore, $\backslash cos\; y\; =\; -\backslash cos\backslash left(\backslash frac\{\backslash pi\}\{3\}\backslash right)$.

Using the property $-\backslash cos(\backslash theta)\; =\; \backslash cos(\backslash pi\; -\; \backslash theta)$, we can write $\backslash cos\; y\; =\; \backslash cos\backslash left(\backslash pi\; -\; \backslash frac\{\backslash pi\}\{3\}\backslash right)\; =\; \backslash cos\backslash left(\backslash frac\{2\backslash pi\}\{3\}\backslash right)$.

The range of the principal value branch for $\backslash cos^\{-1\}(x)$ is $\backslash left[0,\; \backslash pi\backslash right]$. Since $\backslash frac\{2\backslash pi\}\{3\}$ lies within this range, it is the principal value.

Thus, the principal value of $\backslash cos^\{-1\}\backslash left(-\backslash frac\{1\}\{2\}\backslash right)$ is $\backslash frac\{2\backslash pi\}\{3\}$.