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Given below are two statements:

Statement I: cot⁡30∘+1cot⁡30∘−1=2(cos⁡30∘+1)\frac{\cot 30^{\circ}+1}{\cot 30^{\circ}-1}=2\left(\cos 30^{\circ}+1\right)

Statement II : 2sin⁡45∘cos⁡45∘−tan⁡45∘cot⁡45∘=02 \sin 45^{\circ} \cos 45^{\circ}-\tan 45^{\circ} \cot 45^{\circ}=0

Solution

✅ Correct Option: 1

Statement I: cot⁡30°+1cot⁡30°−1=2(cos⁡30°+1)\frac{\cot 30° + 1}{\cot 30° - 1} = 2(\cos 30° + 1)

  1. Left side:
  • cot⁡30°=31\cot 30° = \frac{\sqrt{3}}{1}
    • So 3+13−1\frac{\sqrt{3} + 1}{\sqrt{3} - 1}
    • Rationalize denominator: (3+1)(3+1)(3−1)(3+1)\frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}

= 3+23+13−1\frac{3 + 2\sqrt{3} + 1}{3 - 1}

= 4+232\frac{4 + 2\sqrt{3}}{2}

= 2+32 + \sqrt{3}

  1. Right side:

cos⁡30°=32\cos 30° = \frac{\sqrt{3}}{2}

2(32+1)2(\frac{\sqrt{3}}{2} + 1)

= 3+2\sqrt{3} + 2

= 2+32 + \sqrt{3}

Therefore Statement I is TRUE.

Statement II: 2sin⁡45°cos⁡45°−tan⁡45°cot⁡45°=02\sin 45° \cos 45° - \tan 45° \cot 45° = 0

  1. sin⁡45°=cos⁡45°=12\sin 45° = \cos 45° = \frac{1}{\sqrt{2}}

  2. tan⁡45°=cot⁡45°=1\tan 45° = \cot 45° = 1

  1. Let's substitute:

2(12)(12)−(1)(1)2(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) - (1)(1)

= 2(12)−12(\frac{1}{2}) - 1

= 1−11 - 1

= 00

Therefore Statement II is TRUE.

Hence, both statements are true.

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