Use the formula 2 ^-1 x = ^-1 ((2x)/(1-x2)).

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Here are the evaluations for question 30: i) Evaluate tan (2 tan⁻¹ 1/5 − (π)/(4)) Step 1: Use the formula 2 tan⁻¹ x = tan⁻¹ ((2x)/(1−x²)). Let x = 1/5. 2 tan⁻¹ 1/5 = tan⁻¹ ((2 · 1/5)/(1 − (1/5)²)); 2 tan⁻¹ 1/5 = tan⁻¹ ((2/5)/(1 − 1/25)); 2 tan⁻¹ 1/5 = tan⁻¹ ((2/5)/(24/25)); 2 tan⁻¹ 1/5 = tan⁻¹ (2/5 · 25/24); 2 tan⁻¹ 1/5 = tan⁻¹ (50/120); 2 tan⁻¹ 1/5 = tan⁻¹ 5/12; Step 2: Substitute this back into the expression and use (π)/(4) = tan⁻¹ 1. The expression becomes: tan (tan⁻¹ 5/12 − tan⁻¹ 1) Step 3: Use the formula tan⁻¹ x − tan⁻¹ y = tan⁻¹ ((x−y)/(1+xy)). Let x = 5/12 and y = 1. tan⁻¹ 5/12 − tan⁻¹ 1 = tan⁻¹ ((5/12 − 1)/(1 + 5/12 · 1)); tan⁻¹ 5/12 − tan⁻¹ 1 = tan⁻¹ (((5−12)/(12))/((12+5)/(12))); tan⁻¹ 5/12 − tan⁻¹ 1 = tan⁻¹ ((−7/12)/(17/12)); tan⁻¹ 5/12 − tan⁻¹ 1 = tan⁻¹ (−7/17) Step 4: Evaluate the final expression. tan (tan⁻¹ (−7/17)) = −7/17 The final answer is −7/17. ii) Evaluate tan (1/2 cos⁻¹ (√(5))/(3)) Step 1: Let θ = cos⁻¹ (√(5))/(3). This implies cos θ = (√(5))/(3). We need to find tan ((θ)/(2)). Step 2: Use the half-angle formula tan (θ)/(2) = (1 − cos θ)/(sin θ). First, find sin θ. Since θ = cos⁻¹ (√(5))/(3), θ is in the first quadrant, so sin θ > 0. sin² θ = 1 − cos² θ; sin² θ = 1 − ((√(5))/(3))²; sin² θ = 1 − 5/9; sin² θ = (9−5)/(9); sin² θ = 4/9; sin θ = √(4/9) = 2/3 Step 3: Substitute cos θ and sin θ into the half-angle formula. tan (θ)/(2) = (1 − (√(5))/(3))/(2/3); tan (θ)/(2) = ((3 − √(5))/(3))/(2/3); tan (θ)/(2) = (3 − √(5))/(2) The final answer is (3 − √(5))/(2).

Use the formula 2 ^-1 x = ^-1 ((2x)/(1-x2)).

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Step-by-step solution

Part i· 4 steps

Use the formula $2 \tan^{-1} x = \tan^{-1} \left(\frac{2x}{1-x^2}\right)$ .

Substitute this back into the expression and use $\frac{\pi}{4} = \tan^{-1} 1$ .

Use the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\frac{x-y}{1+xy}\right)$ .

Evaluate the final expression.

Part ii· 3 steps
$\frac{3 - \sqrt{5}}{2}$

Let $\theta = \cos^{-1} \frac{\sqrt{5}}{3}$ .

Use the half-angle formula $\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}$ .

Substitute $\cos \theta$ and $\sin \theta$ into the half-angle formula.

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Use the formula 2 ^-1 x = ^-1 ((2x)/(1-x2)).

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Step-by-step solution

Part i· 4 steps

Use the formula 2tan⁡−1x=tan⁡−1(2x1−x2)2 \tan^{-1} x = \tan^{-1} \left(\frac{2x}{1-x^2}\right)2tan−1x=tan−1(1−x22x​).

Substitute this back into the expression and use π4=tan⁡−11\frac{\pi}{4} = \tan^{-1} 14π​=tan−11.

Use the formula tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy)\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\frac{x-y}{1+xy}\right)tan−1x−tan−1y=tan−1(1+xyx−y​).

Evaluate the final expression.

Part ii· 3 steps3−52\frac{3 - \sqrt{5}}{2}23−5​​

Let θ=cos⁡−153\theta = \cos^{-1} \frac{\sqrt{5}}{3}θ=cos−135​​.

Use the half-angle formula tan⁡θ2=1−cos⁡θsin⁡θ\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}tan2θ​=sinθ1−cosθ​.

Substitute cos⁡θ\cos \thetacosθ and sin⁡θ\sin \thetasinθ into the half-angle formula.

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Use the formula $2 \tan^{-1} x = \tan^{-1} \left(\frac{2x}{1-x^2}\right)$ .

Substitute this back into the expression and use $\frac{\pi}{4} = \tan^{-1} 1$ .

Use the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\frac{x-y}{1+xy}\right)$ .

Part ii· 3 steps
$\frac{3 - \sqrt{5}}{2}$

Let $\theta = \cos^{-1} \frac{\sqrt{5}}{3}$ .

Use the half-angle formula $\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}$ .

Substitute $\cos \theta$ and $\sin \theta$ into the half-angle formula.