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Question Number 117852 by bemath last updated on 14/Oct/20

$$\mathrm{Determine}\mathrm{the}\mathrm{value}\mathrm{of}$$$$\left(1\right)(\tan \frac{7\pi }{24}+\tan \frac{5\pi }{24}).\cos \frac{\pi }{12}+2.$$$$\left(2\right)\sqrt[4]{\frac{9-4\sqrt{5}}{5\mathrm{x}}}.{(5\sqrt{\mathrm{x}}+\sqrt{20\mathrm{x}})}^{0.5}.2^{-1}=?$$

Answered by bobhans last updated on 14/Oct/20

$$\left(2\right)\sqrt[4]{\frac{9-2\sqrt{20}}{5\mathrm{x}}}=\sqrt[4]{\frac{{(\sqrt{5}-2)}^2}{5\mathrm{x}}}=\frac{\sqrt{\sqrt{5}-2}}{\sqrt[4]{5\mathrm{x}}}$$$$\Rightarrow \sqrt[4]{\frac{9-4\sqrt{5}}{5\mathrm{x}}}.{(5\sqrt{\mathrm{x}}+\sqrt{20\mathrm{x}})}^{0.5}.2^{-1}=$$$$\Rightarrow \frac{\sqrt{\sqrt{5}-2}}{\sqrt[4]{5\mathrm{x}}}.\sqrt{5\sqrt{\mathrm{x}}+2\sqrt{5\mathrm{x}}}.\frac{1}{2}=$$$$\Rightarrow \frac{\sqrt{(\sqrt{5}-2)(5\sqrt{\mathrm{x}}+2\sqrt{5\mathrm{x}})}}{2\sqrt[4]{5\mathrm{x}}}=\frac{\sqrt{5\sqrt{5\mathrm{x}}+10\sqrt{\mathrm{x}}-10\sqrt{\mathrm{x}}-4\sqrt{5\mathrm{x}}}}{2\sqrt[4]{5\mathrm{x}}}$$$$=\frac{\sqrt{\sqrt{5\mathrm{x}}}}{2\sqrt[4]{5\mathrm{x}}}=\frac{1}{2}$$

Answered by bobhans last updated on 14/Oct/20

$$\left(1\right)\mathrm{letting}\mathrm{x}=\frac{\pi }{24}$$$$\Rightarrow (\tan 7\mathrm{x}+\tan 5\mathrm{x}).\cos 2\mathrm{x}+2=$$$$\Rightarrow \left(\frac{\sin 7\mathrm{x}.\cos 5\mathrm{x}+\cos 7\mathrm{xsin}5\mathrm{x}}{\cos 7\mathrm{x}\cos 5\mathrm{x}}\right).\cos 2\mathrm{x}+2=$$$$\Rightarrow \frac{\sin 12\mathrm{x}}{2\cos 7\mathrm{x}\cos 5\mathrm{x}}.2\cos 2\mathrm{x}+2=$$$$\Rightarrow \frac{2\cos 2\mathrm{x}}{\cos 12\mathrm{x}+\cos 2\mathrm{x}}+2;[\sin 12\mathrm{x}=\sin \frac{\pi }{2}=1]$$$$[\mathrm{\&}\cos 12\mathrm{x}=\cos \frac{\pi }{2}=0]$$$$\Rightarrow \frac{2\cos \frac{\pi }{12}}{\cos \frac{\pi }{12}}+2=4$$

Answered by mindispower last updated on 14/Oct/20

$$tg\left(a\right)+tg\left(b\right)=\frac{sin(a+b)}{cos\left(a\right)cos\left(b\right)}=\frac{2sin(a+b)}{cos(a-b)+cos(a+b)}$$$$\Rightarrow tg\left(\frac{7\pi }{24}\right)+tg\left(\frac{5\pi }{24}\right)=\frac{2sin\left(\frac{\pi }{2}\right)}{cos\left(\frac{\pi }{12}\right)+cos\left(\frac{\pi }{2}\right)}$$$$\frac{2}{cos\left(\frac{\pi }{12}\right)}$$$$1\iff \left(\frac{2}{cos\left(\frac{\pi }{12}\right)}\right)cos\left(\frac{\pi }{12}\right)+2=4$$

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