# Prove that

Question:

Prove that

(i) $\cos 15^{\circ}-\sin 15^{\circ}=\frac{1}{\sqrt{2}}$

(ii) $\cot 105^{\circ}-\tan 105^{\circ}=2 \sqrt{3}$

(iii) $\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}=-1$

Solution:

(i) $\cos 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$

$\sin 15^{0}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$

$\cos 15^{0}-\sin 15^{0}=\frac{\sqrt{3}+1}{2 \sqrt{2}}-\frac{\sqrt{3}-1}{2 \sqrt{2}}$

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

$=\frac{2}{2 \sqrt{2}}$

$=\frac{1}{\sqrt{2}}$

(ii) $\cot 105^{\circ}-\tan 105^{\circ}=\cot \left(180^{\circ}-75^{\circ}\right)-\tan \left(180^{\circ}-75^{\circ}\right)$

(II quadrant tanx is negative and cotx as well)

$=-\cot 75^{\circ}-\left(-\tan 75^{\circ}\right)$

$=\tan 75^{\circ}-\cot 75^{\circ}$

$\operatorname{Tan} 75^{\circ}=\frac{\sin 75^{\circ}}{\cos 75^{\circ}} \Rightarrow \frac{\sin \left(90^{\circ}-15^{\circ}\right)}{\cos \left(90^{\circ}-15^{\circ}\right)}=\frac{-\cos 15^{\circ}}{\sin 15^{\circ}}$

(using $\sin \left(90^{\circ}-x\right)=-\cos x$ and $\cos \left(90^{\circ}-x\right)=\sin x$ )

$=-\frac{\frac{\sqrt{3}+1}{2 \sqrt{2}}}{\frac{\sqrt{3}-1}{2 \sqrt{2}}} \Rightarrow \frac{-\sqrt{3}-1}{\sqrt{3}-1}$

Cot75° $=\frac{1}{\tan 75^{\circ}} \Rightarrow \frac{\sqrt{3}-1}{-\sqrt{3}-1}$

$\operatorname{Cot} 105^{\circ}-\tan 105^{\circ}$

$=\frac{\sqrt{3}-1}{-\sqrt{3}-1}-\frac{-\sqrt{3}-1}{\sqrt{3}-1} \Rightarrow \frac{(\sqrt{3}-1)-(-\sqrt{3}-1)}{(-\sqrt{3}-1)(\sqrt{3}-1)}=\frac{3+1-2 \sqrt{3}-(3+1+2 \sqrt{3})}{(-3+1-\sqrt{3}+\sqrt{3})}$

$=\frac{-4 \sqrt{3}}{-2} \Rightarrow 2 \sqrt{3}$

(iii) $\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \cdot \tan 66^{\circ}}=\tan \left(69^{\circ}+66^{\circ}\right) \Rightarrow \tan 135^{\circ}=\tan \left(180^{\circ}-45^{\circ}\right)$

$\Rightarrow-\tan 45^{\circ}=-1$