Question

proove that

tan 82×1/2 = (√3+√2)(√2+√1)=√2+√3+√4+√6​

Answer 1

The proof is given below:

$\tan 82.5^\circ$

$=\tan(60^\circ+22.5^\circ)$

$=\frac{\tan 60^\circ+\tan 22.5^\circ}{1-\tan 60^\circ\tan 22.5^\circ}$

$=\frac{\sqrt{3}+\tan 22.5^\circ}{1-\sqrt{3}\tan 22.5^\circ}$

Now,

$\tan (2\times 22.5^\circ)=\frac{2\tan 22.5^\circ}{1-\tan^2 22.5^\circ}$

Let $\tan 22.5^\circ = x$

Thus,

$\tan 45^\circ=\frac{2x}{1-x^2}$

$\implies 1=\frac{2x}{1-x^2}$

$\implies 1-x^2=2x$

$\implies x^2+2x=1$

$\implies x^2+2x+1=2$

$\implies (x+1)^2=2$

$\implies x+1=\pm\sqrt{2}$

$\implies \tan 22.5^\circ=-1\pm\sqrt{2}$

But the value of $\tan 22.5^\circ$ is positive

Therefore,

$\tan 22.5^\circ=-1+\sqrt{2}$

Thus,

$\tan 82.5^\circ=\frac{\sqrt{3}+(\sqrt{2}-1)}{1-\sqrt{3}(\sqrt{2}-1)}$

$\implies \tan 82.5^\circ=\frac{\sqrt{3}+\sqrt{2}-1}{1-(\sqrt{6}-\sqrt{3})}$

$\implies \tan 82.5^\circ=\frac{(\sqrt{3}+\sqrt{2}-1)(1+\sqrt{6}-\sqrt{3})}{(1-(\sqrt{6}-\sqrt{3}))(1+\sqrt{6}-\sqrt{3})}$

$\implies \tan 82.5^\circ=\frac{(\sqrt{3}+\sqrt{2}-1+\sqrt{18}+\sqrt{12}-\sqrt{6}-3-\sqrt{6}+\sqrt{3})}{((1-(6+3-2\sqrt{18})}$

$\implies \tan 82.5^\circ=\frac{(\sqrt{3}+\sqrt{2}-1+3\sqrt{2}+2\sqrt{3}-2\sqrt{6}-3-\sqrt{3})}{(-8+6\sqrt{3})}$

$\implies \tan 82.5^\circ=\frac{(4\sqrt{3}+4\sqrt{2}-4-2\sqrt{6})}{-8+6\sqrt{2}}$

$\implies \tan 82.5^\circ=\frac{(2\sqrt{3}+2\sqrt{2}-2-\sqrt{6})}{-4+3\sqrt{2}}$

$\implies \tan 82.5^\circ=\frac{(2\sqrt{3}+2\sqrt{2}-2-\sqrt{6})(4+3\sqrt{2})}{(-4+3\sqrt{2})(4+3\sqrt{2})}$

$\implies \tan 82.5^\circ=\frac{8\sqrt{3}+8\sqrt{2}-8-4\sqrt{6}+6\sqrt{6}+12-6\sqrt{2}-6\sqrt{3}}{18-16}$

$\implies \tan 82.5^\circ=\frac{8\sqrt{3}+8\sqrt{2}-8-4\sqrt{6}+6\sqrt{6}+12-6\sqrt{2}-6\sqrt{3}}{18-16}$

$\implies \tan 82.5^\circ=\frac{2\sqrt{3}+2\sqrt{2}+2\sqrt{6}+4}{2}$

$\implies \tan 82.5^\circ=\sqrt{3}+\sqrt{2}+\sqrt{6}+2$

$\implies \boxed{\tan 82.5^\circ=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}}$

$\implies\tan82.5^\circ=\sqrt{2}+\sqrt{3}+\sqrt{2}\times\sqrt{2}+\sqrt{2}\times\sqrt{3}$

$\implies\tan82.5^\circ=1(\sqrt{2}+\sqrt{3})+\sqrt{2}(\sqrt{2}+\sqrt{3})$

$\implies\tan82.5^\circ=(\sqrt{2}+\sqrt{3})(1+\sqrt{2})$

$\implies\boxed{\tan82.5^\circ=(\sqrt{3}+\sqrt{2})(\sqrt{2}+\sqrt{1})}$

Hope this answer is helpful.

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