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Write the given expression as an algebraic expression in xx.\newlinetan⁡(2cos⁡−1(x))\tan(2\cos^{-1}(x))

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Full solution

Q. Write the given expression as an algebraic expression in xx.\newlinetan⁡(2cos⁡−1(x))\tan(2\cos^{-1}(x))
  1. Use Double-Angle Formula: We need to express tan⁡(2cos⁡−1(x))\tan(2\cos^{-1}(x)) as an algebraic expression in xx. To do this, we will use the double-angle formula for tangent and the trigonometric identity that relates the cosine of an angle to the tangent of half that angle.
  2. Find Tangent of θ\theta: The double-angle formula for tangent is tan⁡(2θ)=2tan⁡(θ)1−tan⁡2(θ)\tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)}. We will let θ=cos⁡−1(x)\theta = \cos^{-1}(x), so we need to find an expression for tan⁡(cos⁡−1(x))\tan(\cos^{-1}(x)).
  3. Apply Pythagorean Theorem: To find tan⁡(cos⁡−1(x))\tan(\cos^{-1}(x)), we consider a right triangle where the angle θ\theta has a cosine of xx. By definition of cosine, cos⁡(θ)=adjacenthypotenuse\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}. If we let the adjacent side be xx and the hypotenuse be 11, then we need to find the opposite side using the Pythagorean theorem.
  4. Calculate Opposite Side: The Pythagorean theorem states that in a right triangle with sides aa, bb, and hypotenuse cc, a2+b2=c2a^2 + b^2 = c^2. Here, we have the adjacent side a=xa = x and hypotenuse c=1c = 1, so we need to solve for the opposite side bb: b2=c2−a2=12−x2=1−x2b^2 = c^2 - a^2 = 1^2 - x^2 = 1 - x^2.
  5. Determine Tangent of θ\theta: Taking the square root of both sides to find bb, we get b=1−x2b = \sqrt{1 - x^2}. We must be careful to consider only the positive square root since we are dealing with lengths in a triangle.
  6. Substitute into Formula: Now we have the opposite side b=1−x2b = \sqrt{1 - x^2} and the adjacent side a=xa = x. The tangent of θ\theta is tan⁡(θ)=oppositeadjacent=1−x2x\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{1 - x^2}}{x}.
  7. Square Tangent of θ:\theta: Substituting tan⁡(θ)\tan(\theta) into the double-angle formula, we get tan⁡(2θ)=2tan⁡(θ)1−tan⁡2(θ)=2(1−x2/x)1−(1−x2/x)2\tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} = \frac{2(\sqrt{1 - x^2} / x)}{1 - (\sqrt{1 - x^2} / x)^2}.
  8. Simplify Expression: Simplify the expression by squaring tan⁡(θ)\tan(\theta): (1−x2x)2=1−x2x2\left(\frac{\sqrt{1 - x^2}}{x}\right)^2 = \frac{1 - x^2}{x^2}.
  9. Combine Denominator Terms: Substitute this back into the double-angle formula: tan⁡(2θ)=2(1−x2)x/(1−1−x2x2)\tan(2\theta) = \frac{2(\sqrt{1 - x^2})}{x} / \left(1 - \frac{1 - x^2}{x^2}\right).
  10. Final Simplification: Combine the terms in the denominator: 1−1−x2x2=x2−(1−x2)x2=x2−1+x2x2=2x2−1x21 - \frac{1 - x^2}{x^2} = \frac{x^2 - (1 - x^2)}{x^2} = \frac{x^2 - 1 + x^2}{x^2} = \frac{2x^2 - 1}{x^2}.
  11. Final Simplification: Combine the terms in the denominator: 1−1−x2x2=x2−(1−x2)x2=x2−1+x2x2=2x2−1x21 - \frac{1 - x^2}{x^2} = \frac{x^2 - (1 - x^2)}{x^2} = \frac{x^2 - 1 + x^2}{x^2} = \frac{2x^2 - 1}{x^2}.Now we have tan⁡(2θ)=2(1−x2/x)(2x2−1)/x2\tan(2\theta) = \frac{2(\sqrt{1 - x^2} / x)}{(2x^2 - 1) / x^2}. To simplify, multiply both the numerator and the denominator by x2x^2.
  12. Final Simplification: Combine the terms in the denominator: 1−1−x2x2=x2−(1−x2)x2=x2−1+x2x2=2x2−1x21 - \frac{1 - x^2}{x^2} = \frac{x^2 - (1 - x^2)}{x^2} = \frac{x^2 - 1 + x^2}{x^2} = \frac{2x^2 - 1}{x^2}.Now we have tan⁡(2θ)=2(1−x2)/x(2x2−1)/x2\tan(2\theta) = \frac{2(\sqrt{1 - x^2}) / x}{(2x^2 - 1) / x^2}. To simplify, multiply both the numerator and the denominator by x2x^2.After multiplying, we get tan⁡(2θ)=2x1−x22x2−1\tan(2\theta) = \frac{2x \sqrt{1 - x^2}}{2x^2 - 1}.
  13. Final Simplification: Combine the terms in the denominator: 1−1−x2x2=x2−(1−x2)x2=x2−1+x2x2=2x2−1x21 - \frac{1 - x^2}{x^2} = \frac{x^2 - (1 - x^2)}{x^2} = \frac{x^2 - 1 + x^2}{x^2} = \frac{2x^2 - 1}{x^2}. Now we have tan⁡(2θ)=2(1−x2)/x(2x2−1)/x2\tan(2\theta) = \frac{2(\sqrt{1 - x^2}) / x}{(2x^2 - 1) / x^2}. To simplify, multiply both the numerator and the denominator by x2x^2. After multiplying, we get tan⁡(2θ)=2x1−x22x2−1\tan(2\theta) = \frac{2x \sqrt{1 - x^2}}{2x^2 - 1}. This is the simplified algebraic expression for tan⁡(2cos⁡−1(x))\tan(2\cos^{-1}(x)): tan⁡(2cos⁡−1(x))=2x1−x22x2−1\tan(2\cos^{-1}(x)) = \frac{2x \sqrt{1 - x^2}}{2x^2 - 1}.

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