Properties of inverse trigonometric functions

The governing concept is the principal-value identity for inverse tangent and inverse cotangent. For a positive real number x, let θ = tan⁻¹x. Then θ lies in (0, π/2), and cot⁻¹x is the complementary angle π/2 − θ under the standard Class 12 principal-value convention. Therefore tan⁻¹x + cot⁻¹x = θ + (π/2 − θ) = π/2. Hence option B is correct. Option A would occur only for a particular value such as x = 1, not for every positive x. Options C and D ignore the complementary principal-value relationship. The restriction x > 0 also avoids sign-related branch complications.

The governing identity on the principal branch of inverse cosine is cos⁻¹(−x) = π − cos⁻¹x for x ∈ [−1,1]. Let α = cos⁻¹x. Then α lies in [0,π] and cos α = x. Using the cosine identity, cos(π − α) = −cos α = −x. Also, π − α remains in [0,π], the principal range of cos⁻¹. Therefore cos⁻¹(−x) = π − α = π − cos⁻¹x, so option C is correct. Option A incorrectly treats arccos as an odd function; it is not. Option B ignores the sign change, and option D is a sine-related shift that does not generally have cosine −x. The range condition is what makes the identity unambiguous.

If \(x\in\left[-1,1\right]\), then \(\cos(\cos^{-1}x)=x\). Here \(x=-\frac{3}{4}\) lies in the domain.

Here \(\theta\in\left[0,\pi\right]\) and \(\cos\theta=\frac{8}{17}\), so \(\theta\) is in the first quadrant and \(\sin\theta=\frac{15}{17}\). The triangle method is fast.

From \(\tan\theta=\frac{7}{24}\), the hypotenuse is \(\sqrt{7^2+24^2}=25\), so \(\sin\theta=\frac{7}{25}\). Build a triangle from the ratio.

Let \(\theta=\sin^{-1}\frac{4}{5}\), then \(\sin\theta=\frac{4}{5}\) and \(\cos\theta=\frac{3}{5}\). Therefore \(\tan\theta=\frac{4}{3}\).

If \(\theta=\tan^{-1}\frac{12}{5}\), then \(\tan\theta=\frac{12}{5}\) and the hypotenuse is (13). Hence \(\sin\theta=\frac{12}{13}\).

If \(\tan\theta=\frac{15}{8}\), then the hypotenuse is (17). In the principal range \(\theta\) is in the first quadrant, so \(\cos\theta=\frac{8}{17}\).

For every \(x\in\left[-1,1\right]\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Here \(x=\frac{2}{3}\) is valid.

From \(\cot\theta=\frac{3}{4}\), the base is (3), perpendicular is (4), and hypotenuse is (5). Hence \(\cos\theta=\frac{3}{5}\).

By the formula, the sum is \(\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)=\tan^{-1}1=\frac{\pi}{4}\). While adding \(\tan^{-1}\) terms, check \(ab<1\).

Let \(\theta=\cos^{-1}\frac{5}{13}\), then \(\cos\theta=\frac{5}{13}\) and \(\sin\theta=\frac{12}{13}\). Therefore \(\tan\theta=\frac{12}{5}\).

The inverse tangent function is an odd function. For an odd function, changing the sign of the input changes the sign of the output, so \(f(-x)=-f(x)\). Applying this property to the inverse tangent function gives \(\tan^{-1}(-x)=-\tan^{-1}(x)\). Therefore option A is correct.

The principal value of \(\tan^{-1}(x)\) lies in \((-\frac{\pi}{2},\frac{\pi}{2})\), and the negative of a principal value remains in the same interval. This confirms that no extra \(\pi\) or \(\frac{\pi}{2}\) term is needed. Option B would incorrectly ignore the negative input, while options C and D are not the odd-function identity and generally give a different value. The identity follows because tangent itself is odd and its inverse preserves the corresponding sign relationship on its principal domain.

\(\cos^{-1}(-x)=\pi-\cos^{-1}x\) because the output remains in \([0,\pi]\). Do not treat it as an odd function.

If \(\theta=\cos^{-1}x\), then \(\theta\in[0,\pi]\) and \(\sin\theta\ge0\). Hence the value is \(\sqrt{1-x^2}\).

If (\theta=\sin^{-1}x), then (\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]) and (\cos\theta\ge0). So the positive root should be taken.

Let (\theta=\sin^{-1}x), then (\sin\theta=x) and (\cos\theta=\sqrt{1-x^2}). Hence (\tan\theta=\frac{x}{\sqrt{1-x^2}}).

If \(\theta=\tan^{-1}x\), then \(\tan\theta=x\) and the hypotenuse is \(\sqrt{1+x^2}\). So \(\sin\theta=\frac{x}{\sqrt{1+x^2}}\).

Taking \(\theta=\tan^{-1}x\), the adjacent side is (1) and the hypotenuse is \(\sqrt{1+x^2}\). Thus \(\cos\theta=\frac{1}{\sqrt{1+x^2}}\).

This follows directly from \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Here \(x=\frac{\sqrt{3}}{2}\) is in the valid domain.

\(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\), so \(\cos\theta\) is positive. From the (3,4,5) triangle, \(\cos\theta=\frac{4}{5}\).

From \(\tan\theta=\frac{5}{12}\), opposite is (5), adjacent is (12), and hypotenuse is (13). Hence \(\sin\theta=\frac{5}{13}\).

The governing complementary-angle identity is sin⁻¹x + cos⁻¹x = π/2. Rearranging gives cos⁻¹x = π/2 − sin⁻¹x. Both inverse functions must have real values, and the real domain of each is −1 ≤ x ≤ 1. Thus the identity is true precisely for x ∈ [−1, 1], making option A correct. The endpoints are included because sin⁻¹(±1) and cos⁻¹(±1) are defined. Option B is too broad: inverse sine and inverse cosine are not real for every real x. Options C and D lie outside the real domain and therefore cannot satisfy the stated real-valued identity.

\(\cos^{-1}x\) is defined for \(-1\le x\le 1\). On this interval, as \(x\) increases, the principal angle \(\theta=\cos^{-1}x\) decreases (because \(\cos\theta\) decreases over \([0,\pi]\)). Therefore, \(f(x)=\cos^{-1}x\) is a decreasing function on \([-1,1]\).

\(\frac{x}{\sqrt{1+x^2}}\) always lies in ([-1,1]). Therefore the identity is valid for all real (x).