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Given: $\cos \frac{X}{2}=\frac{12}{13}$ and $\mathrm{x}$ lies in Quadrant $\mathrm{I}$ i.e, All the trigonometric ratios are positive in I quadrant

To Find: (i) $\sin x$ ii) $\cos x$ iii) $\cot x$

(i) $\sin x$

Formula used:

We have, $\operatorname{Sin} x=\sqrt{1-\cos ^{2} x}$

We know that, $\cos \frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}(\because \cos x$ is positive in I quadrant)

$\Rightarrow 2 \cos ^{2} \frac{x}{2}-1=\cos x$

$\Rightarrow 2 \times\left(\frac{12}{13}\right)^{2}-1=\cos x$

$\Rightarrow 2 \times\left(\frac{144}{169}\right)-1=\cos x$

$\Rightarrow \cos x=\frac{119}{169}$

Since, $\operatorname{Sin} x=\sqrt{1-\cos ^{2} x}$

$\Rightarrow \operatorname{Sin} x=\sqrt{1-\left(\frac{119}{169}\right)^{2}}$

$\Rightarrow \operatorname{Sin} x=\frac{120}{169}$

Hence, we have $\operatorname{Sin} x=\frac{120}{169}$.

ii) $\cos x$

Formula used:

We know that, $\cos \frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}(\because \cos x$ is positive in I quadrant)

$\Rightarrow 2 \cos ^{2} \frac{x}{2}-1=\cos x$

$\Rightarrow 2 \times\left(\frac{12}{13}\right)^{2}-1=\cos x$

$\Rightarrow 2 \times\left(\frac{144}{169}\right)-1=\cos x$

$\Rightarrow \cos x=\frac{119}{169}$

iii) $\cot x$

Formula used:

$\cot x=\frac{\cos x}{\sin x}$

$\cot x=\frac{\frac{119}{169}}{\frac{120}{169}}=\frac{119}{169} \times \frac{169}{120}=\frac{119}{120}$

Hence, we have $\cot x=\frac{119}{120}$