Simplify tan⁻¹(x/(a + √(a² − x²))) Problem Simplify: \( \tan^{-1}\left(\frac{x}{a + \sqrt{a^2 – x^2}}\right), \quad -a < x < a \) Solution (Substitution Method) Let: \[ x = a \sin \theta \] Then, \[ \sqrt{a^2 – x^2} = a \cos \theta \] So the expression becomes: \[ \tan^{-1}\left(\frac{a \sin \theta}{a + a \cos \theta}\right) = […]
Write in the simplest form tan^-1(x/a+(√a^2-x^2)). -a < x < a Read More »
Simplify tan⁻¹(√(a − x)/√(a + x)) Problem Simplify: \( \tan^{-1}\left(\frac{\sqrt{a – x}}{\sqrt{a + x}}\right), \quad -a < x < a \) Solution (Substitution Method) Let: \[ x = a \cos \theta \] Then, \[ a – x = a(1 – \cos \theta), \quad a + x = a(1 + \cos \theta) \] So, \[ \frac{\sqrt{a
Write in the simplest form tan^-1(√a-x)/(√a+x), -a less than x less than a Read More »
Simplify tan⁻¹((√(1 + x²) + 1)/x) Problem Simplify: \( \tan^{-1}\left(\frac{\sqrt{1 + x^2} + 1}{x}\right), \quad x \ne 0 \) Solution (Substitution Method) Let: \[ x = \tan \theta \] Then, \[ \sqrt{1 + x^2} = \sec \theta \] So the expression becomes: \[ \tan^{-1}\left(\frac{\sec \theta + 1}{\tan \theta}\right) \] Simplify: \[ \frac{\sec \theta + 1}{\tan
Write in the simplest form tan^-1{((√1+x^2) +1)/x}, x≠0 Read More »
Simplify tan⁻¹((√(1 + x²) − 1)/x) Problem Simplify: \( \tan^{-1}\left(\frac{\sqrt{1 + x^2} – 1}{x}\right), \quad x \ne 0 \) Solution (Substitution Method) Let: \[ x = \tan \theta \] Then, \[ \sqrt{1 + x^2} = \sqrt{1 + \tan^2 \theta} = \sec \theta \] So the expression becomes: \[ \tan^{-1}\left(\frac{\sec \theta – 1}{\tan \theta}\right) \] Simplify:
Write in the simplest form tan^-1{((√1+x^2) -1)/x}, x≠0 Read More »
Simplify tan⁻¹(√(1 + x²) − x) by Substitution Problem Simplify: \( \tan^{-1}(\sqrt{1 + x^2} – x), \quad x \in \mathbb{R} \) Solution (Substitution Method) Let: \[ x = \tan \theta \] Then, \[ \sqrt{1 + x^2} = \sqrt{1 + \tan^2 \theta} = \sec \theta \] So the expression becomes: \[ \tan^{-1}(\sec \theta – \tan \theta)
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Simplify tan⁻¹(x + √(1 + x²)) by Long Method Problem Simplify: \( \tan^{-1}(x + \sqrt{1 + x^2}), \quad x \in \mathbb{R} \) Solution (Long Method) Let: \[ \theta = \tan^{-1}(x + \sqrt{1 + x^2}) \] Then, \[ \tan \theta = x + \sqrt{1 + x^2} \] Taking reciprocal, \[ \cot \theta = \frac{1}{x + \sqrt{1
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Simplify cot⁻¹(a/√(x² − a²)) Problem Simplify: \( \cot^{-1}\left(\frac{a}{\sqrt{x^2 – a^2}}\right), \quad |x| > a \) Solution Let: \[ \theta = \cot^{-1}\left(\frac{a}{\sqrt{x^2 – a^2}}\right) \] Then, \[ \cot \theta = \frac{a}{\sqrt{x^2 – a^2}} \] Taking reciprocal: \[ \tan \theta = \frac{\sqrt{x^2 – a^2}}{a} \] Now, construct a right triangle: Opposite = \( \sqrt{x^2 – a^2} \)
Write in the simplest form : cot^-1{a/(√x^2 -a^2)}, where |x| > a. Read More »
Evaluate cot⁻¹{cot(21π/4)} Problem Evaluate: \( \cot^{-1}(\cot \frac{21\pi}{4}) \) Solution First, reduce the angle: \[ \frac{21\pi}{4} = 5\pi + \frac{\pi}{4} \] Since cotangent has period \( \pi \): \[ \cot \frac{21\pi}{4} = \cot \frac{\pi}{4} \] Now, \[ \cot \frac{\pi}{4} = 1 \] Thus the expression becomes: \[ \cot^{-1}(1) \] Recall the principal value range of \(
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Evaluate cot⁻¹{cot(−8π/3)} Problem Evaluate: \( \cot^{-1}(\cot(-\frac{8\pi}{3})) \) Solution First, reduce the angle using periodicity of cotangent (period \( \pi \)): \[ -\frac{8\pi}{3} = -3\pi + \frac{\pi}{3} \] So, \[ \cot\left(-\frac{8\pi}{3}\right) = \cot\left(\frac{\pi}{3}\right) \] Now, \[ \cot \frac{\pi}{3} = \frac{1}{\sqrt{3}} \] Thus the expression becomes: \[ \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) \] Recall the principal value range of \( \cot^{-1}
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Evaluate cot⁻¹(cot 19π/6) Problem Evaluate: \( \cot^{-1}(\cot \frac{19\pi}{6}) \) Solution First, reduce the angle: \[ \frac{19\pi}{6} = 3\pi + \frac{\pi}{6} \] Since cotangent has period \( \pi \): \[ \cot \frac{19\pi}{6} = \cot \frac{\pi}{6} \] Now, \[ \cot \frac{\pi}{6} = \sqrt{3} \] Thus the expression becomes: \[ \cot^{-1}(\sqrt{3}) \] Recall the principal value range of
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