$\,1.\,$ Find the values :
$\,(i)\,\,\sin(-75^{\circ})\\=-\sin 75^{\circ}\\=-\sin(30^{\circ}+45^{\circ})\\=-[\sin 30^{\circ}\cos 45^{\circ}+\cos 30^{\circ}\sin 45^{\circ}]\\=-[\frac 12. \frac{1}{\sqrt2}+\frac{\sqrt3}{2}.\frac{1}{\sqrt2}]\\=-[\frac{1}{2\sqrt2}+\frac{\sqrt3}{2\sqrt2}]\\=-\frac{\sqrt3+1}{2\sqrt2}\quad \text{(ans.)}$
$\,(ii)\,\,\cos 15^{\circ}\\=\cos(45^{\circ}-30^{\circ})\\=\cos 45^{\circ}\cos 30^{\circ}+\sin 45^{\circ}\sin 30^{\circ}\\=\frac{1}{\sqrt2} \times \frac{\sqrt3}{2}+\frac{1}{\sqrt2} \times \frac 12\\=\frac{\sqrt3}{2\sqrt2}+\frac{1}{2\sqrt2}\\=\frac{\sqrt3+1}{2\sqrt2}\quad \text{(ans.)}$
$\,(iii)\,\,\,\tan(-105^{\circ})\\=-\tan 105^{\circ}\\=-\tan(60^{\circ}+45^{\circ})\\=-\frac{\tan 60^{\circ}+\tan 45^{\circ}}{1-\tan 60^{\circ}\tan 45^{\circ}}\\=-\frac{\sqrt3+1}{1-\sqrt3\times 1}\\=\frac{\sqrt3+1}{\sqrt3-1}\\=\frac{(\sqrt3+1)^2}{(\sqrt3-1)(\sqrt3+1)}\\=\frac{(\sqrt3)^2+2\sqrt3+1}{3-1}\\=\frac{4+2\sqrt3}{2}\\=2+\sqrt3\quad \text{(ans.)}$
$\,(iv)\,\,\sec(-75^{\circ})\\=\sec 75^{\circ}\\=\sec(45^{\circ}+30^{\circ})\\=\frac{1}{\cos(45^{\circ}+30^{\circ})}\\=\frac{1}{\cos45^{\circ}\cos30^{\circ}-\sin 45^{\circ}\sin 30^{\circ}}\\=\frac{1}{\frac{1}{\sqrt2}\times \frac{\sqrt3}{2}-\frac{1}{\sqrt2} \times \frac 12}\\=\frac{2\sqrt2}{\sqrt3-1}\\=\frac{2\sqrt2(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}\\=\frac{2\sqrt2(\sqrt3+1)}{3-1}\\=\frac{2\sqrt2(\sqrt3+1)}{2}\\=\sqrt2(\sqrt3+1)\quad \text{(ans.)}$
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$\,2.\,$ Prove that ,
$\,(i)\,\,\tan 75^{\circ}-\cot 75^{\circ}=4\sin 60^{\circ}$
Sol. L.H.S. $=\tan 75^{\circ}-\cot 75^{\circ}\\=\tan 75^{\circ}-\tan(90^{\circ}-75^{\circ})\\=\tan 75^{\circ}-\cot 15^{\circ}\\=\tan (45^{\circ}+30^{\circ})-\tan(45^{\circ}-30^{\circ})\\=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ}\tan 30^{\circ}}-\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}\tan 30^{\circ}} \\=\frac{1+\frac{1}{\sqrt3}}{1-1.\frac{1}{\sqrt3}}-\frac{1-\frac{1}{\sqrt3}}{1+1.\frac{1}{\sqrt3}}\\=\frac{\sqrt3+1}{\sqrt3-1}-\frac{\sqrt3-1}{\sqrt3+1}\\=\frac{(\sqrt3+1)^2-(\sqrt3-1)^2}{(\sqrt3-1)(\sqrt3+1)}\\=\frac{4 \times \sqrt3\times 1}{(\sqrt3)^2-1^2}\\=\frac{4\sqrt3}{3-1}\\=\frac{4\sqrt3}{2}\\=4 \times \frac{\sqrt3}{2}\\=4\sin 60^{\circ}=\text{ R.H.S.}\,\,\text{(proved)}$
$\,\,2(ii)\,\,\cos\alpha+\cos(120^{\circ}+\alpha)+\cos(120^{\circ}-\alpha)=0$
Sol. L.H.S. $=\cos\alpha+\cos(120^{\circ}+\alpha)+\cos(120^{\circ}-\alpha)\\=\cos\alpha+\cos[90^{\circ}+(\alpha+30^{\circ})]\\+\cos[90^{\circ}-(\alpha-30^{\circ})]\\=\cos\alpha-\sin(\alpha+30^{\circ})+\sin(\alpha-30^{\circ})\\=\cos\alpha-(\sin\alpha\cos30^{\circ}+\cos\alpha\sin 30^{\circ})\\+(\sin\alpha\cos30^{\circ}-\cos\alpha\sin30^{\circ})\\=\cos\alpha-\sin\alpha\cos30^{\circ}-\cos\alpha\sin30^{\circ}\\+\sin\alpha\cos30^{\circ}-\cos\alpha\sin30^{\circ}\\=\cos\alpha-2\cos\alpha\times \frac 12\\=\cos\alpha-\cos\alpha\\=0=\text{R.H.S.(proved)}$
$\,\,2(iii)\,\,\tan 35^{\circ}+\tan 10^{\circ}+\tan 35^{\circ}\tan10^{\circ}\\=1$
Sol. We know, $\,\,\tan 45^{\circ}=1 \\ \Rightarrow \tan(35^{\circ}+10^{\circ})=1 \\ \Rightarrow \frac{\tan 35^{\circ}+\tan 10^{\circ}}{1-\tan 35^{\circ}\tan 10^{\circ}}=1 \\ \Rightarrow \tan 30^{\circ}+\tan 10^{\circ}=1-\tan 35^{\circ}\tan 10^{\circ} \\ \Rightarrow \tan 35^{\circ}+\tan 10^{\circ}+\tan 35^{\circ}\tan 10^{\circ}=1\\ \text{(proved)}$
$\,\,2(iv)\,\,\tan 8\alpha-\tan5\alpha-\tan3\alpha=\tan8\alpha\tan5\alpha\tan3\alpha$
Sol. $\,\,\tan 8\alpha\\=\tan(5\alpha+3\alpha)\\=\frac{\tan5\alpha+\tan3\alpha}{1-\tan5\alpha\tan3\alpha}\\ \Rightarrow \tan8\alpha(1-\tan5\alpha\tan3\alpha)\\=\tan5\alpha+\tan3\alpha \\ \Rightarrow \tan8\alpha-\tan8\alpha\tan5\alpha\tan3\alpha\\=\tan5\alpha+\tan3\alpha \\ \Rightarrow \tan 8\alpha-\tan5\alpha-\tan3\alpha\\=\tan8\alpha\tan5\alpha\tan3\alpha\,\,\text{(proved)}$
$\,\,2(v)\,\,\tan 43^{\circ}=\frac{\cos2^{\circ}-\sin2^{\circ}}{\cos2^{\circ}+\sin2^{\circ}}$
Sol. $\,\,\text{R.H.S.}\\=\frac{\cos2^{\circ}-\sin2^{\circ}}{\cos2^{\circ}+\sin2^{\circ}}\\=\frac{1-\tan2^{\circ}}{1+\tan2^{\circ}}\,\,[**]\\=\frac{\tan45^{\circ}-\tan2^{\circ}}{1+\tan45^{\circ}\tan2^{\circ}}\\=\tan(45^{\circ}-2^{\circ})\\=\tan43^{\circ}\\=\text{L.H.S.(proved)}$
Note [**] : Dividing numerator and denominator by $\,\,\cos2^{\circ}$
$\,\,2(vi)\,\,$ Prove that, $\,\,\tan62^{\circ}=2\tan34^{\circ}+\tan28^{\circ}$
Sol. $\,\,\tan 62^{\circ}\\=\tan(34^{\circ}+28^{\circ})\\=\frac{\tan34^{\circ}+\tan28^{\circ}}{1-\tan34^{\circ}\tan28^{\circ}}\\ \Rightarrow \tan 62^{\circ}(1-\tan 34^{\circ}\tan 28^{\circ})\\=\tan 34^{\circ}+\tan28^{\circ} \\ \Rightarrow \tan 62^{\circ}-\tan 62^{\circ}\tan 28^{\circ}.\tan 34^{\circ}\\=\tan 34^{\circ}+\tan 28^{\circ} \\ \Rightarrow \tan 62^{\circ}-\cot(90^{\circ}-62^{\circ}). \tan 28^{\circ} \\ \times \tan 34^{\circ}=\tan 34^{\circ}+\tan 28^{\circ} \\ \Rightarrow \tan 62^{\circ}-(\cot 28^{\circ}.\tan 28^{\circ})\tan 34^{\circ}\\=\tan 34^{\circ}+\tan 28^{\circ} \\ \Rightarrow \tan 62^{\circ}-\tan 34^{\circ}=\tan 34^{\circ}+\tan 28^{\circ} \\ \Rightarrow \tan62^{\circ}=2\tan34^{\circ}+\tan28^{\circ}\quad \text{(proved)}$
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