April 2026

Evaluate tan(cos⁻¹(8/17)) Problem Evaluate: \( \tan\left(\cos^{-1}\left(\frac{8}{17}\right)\right) \) Solution Let \( \theta = \cos^{-1}\left(\frac{8}{17}\right) \) Then: \[ \cos \theta = \frac{8}{17} \] Construct a right triangle: Adjacent = 8 Hypotenuse = 17 Opposite: \[ \sqrt{17^2 – 8^2} = \sqrt{289 – 64} = \sqrt{225} = 15 \] Now, \[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{15}{8} \] Therefore: […]

Evaluate sec(sin⁻¹(12/13)) Problem Evaluate: \( \sec\left(\sin^{-1}\left(\frac{12}{13}\right)\right) \) Solution Let \( \theta = \sin^{-1}\left(\frac{12}{13}\right) \) Then: \[ \sin \theta = \frac{12}{13} \] Construct a right triangle: Opposite = 12 Hypotenuse = 13 Adjacent: \[ \sqrt{13^2 – 12^2} = \sqrt{169 – 144} = \sqrt{25} = 5 \] Now, \[ \cos \theta = \frac{5}{13} \] \[ \sec \theta

Evaluate cosec(cos⁻¹(3/5)) Problem Evaluate: \( \csc\left(\cos^{-1}\left(\frac{3}{5}\right)\right) \) Solution Let \( \theta = \cos^{-1}\left(\frac{3}{5}\right) \) Then: \[ \cos \theta = \frac{3}{5} \] Using identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] \[ \sin \theta = \sqrt{1 – \cos^2 \theta} = \sqrt{1 – \left(\frac{3}{5}\right)^2} = \sqrt{1 – \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \] Now, \[

Evaluate sin(sec⁻¹(17/8)) Problem Evaluate: \( \sin\left(\sec^{-1}\left(\frac{17}{8}\right)\right) \) Solution Let \( \theta = \sec^{-1}\left(\frac{17}{8}\right) \) Then: \[ \sec \theta = \frac{17}{8} \] So, \[ \cos \theta = \frac{8}{17} \] Using identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] \[ \sin \theta = \sqrt{1 – \cos^2 \theta} = \sqrt{1 – \left(\frac{8}{17}\right)^2} = \sqrt{1 – \frac{64}{289}}

Evaluate sin(tan⁻¹(24/7)) Problem Evaluate: \( \sin\left(\tan^{-1}\left(\frac{24}{7}\right)\right) \) Solution Let \( \theta = \tan^{-1}\left(\frac{24}{7}\right) \) Then: \[ \tan \theta = \frac{24}{7} \] Construct a right triangle: Opposite = 24 Adjacent = 7 Hypotenuse: \[ \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \] Now, \[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{24}{25} \] Therefore:

Evaluate sin(sin⁻¹(7/25)) Problem Evaluate: \( \sin\left(\sin^{-1}\left(\frac{7}{25}\right)\right) \) Solution We use the basic identity: \[ \sin(\sin^{-1}x) = x \] Applying this identity: \[ \sin\left(\sin^{-1}\left(\frac{7}{25}\right)\right) = \frac{7}{25} \] Final Answer \[ \boxed{\frac{7}{25}} \] Explanation The sine function and its inverse cancel each other when applied in this form, provided the value lies within the domain of the

Evaluate sin(cos⁻¹(5/13)) Problem Evaluate: \( \sin\left(\cos^{-1}\left(\frac{5}{13}\right)\right) \) Solution Let \( \theta = \cos^{-1}\left(\frac{5}{13}\right) \) Then: \[ \cos \theta = \frac{5}{13} \] Using identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] \[ \sin \theta = \sqrt{1 – \cos^2 \theta} \] \[ \sin \theta = \sqrt{1 – \left(\frac{5}{13}\right)^2} = \sqrt{1 – \frac{25}{169}} = \sqrt{\frac{144}{169}} =

Simplify sin{2tan⁻¹(√(1−x)/√(1+x))} Problem Simplify: \( \sin\left(2\tan^{-1}\left(\frac{\sqrt{1-x}}{\sqrt{1+x}}\right)\right) \) Solution (Substitution Method) Let: \[ \theta = \tan^{-1}\left(\frac{\sqrt{1-x}}{\sqrt{1+x}}\right) \] Then, \[ \tan \theta = \frac{\sqrt{1-x}}{\sqrt{1+x}} \] Using identity: \[ \sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta} \] Compute: \[ \tan^2 \theta = \frac{1-x}{1+x} \] \[ 1 + \tan^2 \theta = \frac{1+x + 1-x}{1+x} = \frac{2}{1+x} \] Thus,

Simplify sin⁻¹((√(1+x) + √(1−x))/2) Problem Simplify: \( \sin^{-1}\left(\frac{\sqrt{1+x} + \sqrt{1-x}}{2}\right), \quad 0 < x < 1 \) Solution (Substitution Method) Let: \[ x = \cos 2\theta \] Then, \[ 1 + x = 1 + \cos 2\theta = 2\cos^2 \theta \] \[ 1 – x = 1 – \cos 2\theta = 2\sin^2 \theta \] So,

Simplify sin⁻¹((x + √(1 − x²))/√2) Problem Simplify: \( \sin^{-1}\left(\frac{x + \sqrt{1 – x^2}}{\sqrt{2}}\right), \quad -\frac{1}{2} < x < \frac{1}{\sqrt{2}} \) Solution (Substitution Method) Let: \[ x = \sin \theta \] Then, \[ \sqrt{1 – x^2} = \cos \theta \] So the expression becomes: \[ \sin^{-1}\left(\frac{\sin \theta + \cos \theta}{\sqrt{2}}\right) \] Using identity: \[ \sin