JIPMAT 2024 (QA) - Given below are two statements: Statement I: 30^+1 30^-1=2( 30^+1) Statement II : 2 45^ 45^- 45^ 45^=0 | PYQs + Solutions

JIPMAT 2024

Geometry

Trigonometry

Hard

Given below are two statements:
Statement I: $\frac{\cot 30^{\circ}+1}{\cot 30^{\circ}-1}=2\left(\cos 30^{\circ}+1\right)$
Statement II : $2 \sin 45^{\circ} \cos 45^{\circ}-\tan 45^{\circ} \cot 45^{\circ}=0$

Both Statement I and Statement II are true

Both Statement I and Statement II are false

Statement I is true but Statement II is false

Statement I is false but Statement II is true

Correct Option: 1

Statement I:

\frac{\cot 30° + 1}{\cot 30° - 1} = 2(\cos 30° + 1)

1) Left side:

\newline

\cot 30° = \frac{\sqrt{3}}{1}

\newline

- So

\frac{\sqrt{3} + 1}{\sqrt{3} - 1}

\newline

- Rationalize denominator:

\frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}

\frac{3 + 2\sqrt{3} + 1}{3 - 1}

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

2 + \sqrt{3}

2) Right side:

\newline

\cos 30° = \frac{\sqrt{3}}{2}

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

\sqrt{3} + 2

2 + \sqrt{3}

Therefore Statement I is TRUE.

Statement II:

2\sin 45° \cos 45° - \tan 45° \cot 45° = 0

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

\newline

\tan 45° = \cot 45° = 1

3) Let's substitute:

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

2(\frac{1}{2}) - 1

1 - 1

\newline

0

Therefore Statement II is TRUE.

Hence, both statements are true.