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Trigonometric Ratios Of Compounds — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compounds4 Marks
Question
Prove that: $\tan82\frac{1^\circ}{2}=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$
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Answer

$\tan82\frac{1^\circ}{2}=\tan\Big(90-7\frac{1}{7}\Big)$ $=\cot7\frac{1^\circ}{2}$ $=\cot\text{A}$ if $\text{A}=7\frac{1^\circ}{2}$ Now, $\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}}$ $=\frac{2\cos^2\text{x}}{1\sin\text{x}\cos\text{x}}$ $=\frac{1+\cos^2\text{x}}{\sin^2\text{x}}$ $\cot\text{x}=\frac{1+\cos15}{\sin15}$ $=\frac{1+\cos(45-30)}{\sin15}$ $\frac{1+\Big(\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times\frac{1}{2}\Big)}{\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times\frac{}{}\frac{1}{2}}$ $=\frac{2\sqrt{2}+(\sqrt{3}+1)}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$ $=\frac{2\sqrt{2}(\sqrt{3}+1)+(\sqrt{3}+1)^2}{3-1}$ $=\frac{2\sqrt{6}+2\sqrt{2}+4+2\sqrt{3}}{2}$ $\cot\text{x}=\sqrt{6}+\sqrt{2}+2+\sqrt{3}\ .....(1)$ $=\sqrt{2}+2\sqrt{6}+\sqrt{3}$ $=\sqrt{2}(1+\sqrt{2})+\sqrt{3}(\sqrt{2}+1)$ $\cot\text{x}=(\sqrt{2}+1)(\sqrt{2}+\sqrt{3})]\ .....(2)$ From equation (1) and (2) $\tan82\frac{1^\circ}{2}=\cot7\frac{1^\circ}{2}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$ $=(\sqrt{2}+1)(\sqrt{2}+\sqrt{3})$

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