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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 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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