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Question Number 177964 by mathlove last updated on 11/Oct/22

$$\frac{1}{sec15sin15cos30}=?$$

Commented by cortano1 last updated on 11/Oct/22

$$\frac{\cos 15°}{\frac{1}{2}\sqrt{3}\sin 15°}=\frac{2}{\sqrt{3}}\cot 15°=\frac{2}{\sqrt{3}}\times (2+\sqrt{3})$$$$=\frac{4+2\sqrt{3}}{\sqrt{3}}=2+\frac{4\sqrt{3}}{3}$$$$(\bullet )\tan 15°=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$$$$\Rightarrow \cot 15°=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{1}{2}(4+2\sqrt{3})=2+\sqrt{3}$$$$$$

Answered by Ar Brandon last updated on 11/Oct/22

$$\frac{1}{\sec 15°\sin 15°\cos 30°}=\frac{\cos 15°}{\sin 15°}\times \frac{1}{\cos 30°}=\frac{2}{\sqrt{3}\tan 15°}$$$$\cos 2x={2\cos }^{2}x-1=\frac{2}{{\sec }^{2}x}-1=\frac{2}{1+{\tan }^{2}x}-1$$$$\Rightarrow \cos 30°=\frac{2}{1+{\tan }^{2}15°}-1,x=15°$$$$\Rightarrow \frac{\sqrt{3}}{2}+1=\frac{2}{1+{\tan }^{2}15°}\Rightarrow \frac{\sqrt{3}+2}{2}=\frac{2}{1+{\tan }^{2}15°}$$$$\Rightarrow 1+{\tan }^{2}15°=\frac{4}{\sqrt{3}+2}=4(2-\sqrt{3})$$$$\Rightarrow \tan 15°=\sqrt{7-4\sqrt{3}}=\sqrt{{(\sqrt{3}-\sqrt{4})}^2}=\sqrt{4}-\sqrt{3}$$$$\Rightarrow \frac{1}{\sec 15°\sin 15°\cos 30°}=\frac{2}{\sqrt{3}(\sqrt{4}-\sqrt{3})}=\frac{2(2+\sqrt{3})}{\sqrt{3}}$$

Commented by mathlove last updated on 12/Oct/22

$$thanks$$

Answered by Rasheed.Sindhi last updated on 12/Oct/22

$$\frac{1}{\sec 15\sin 15\cos 30}=?$$$$=\frac{1}{\sec 15\sin 15\underset{―}{\underset{\text{blacktrinagledown}}{\sin 60}}}$$$$\sin 60=2\sin 30\cos 30=2\left(2\sin 15\cos 15\right)\cos 30$$$$=4\sin 15\cos 15\cos 30$$$$=\frac{1}{\cancel{\sec 15sin15}\cdot 4\sin 15\cancel{\cos 15cos30}}$$$$=\frac{1}{4\underset{―}{\underset{\text{blacktrinagledown}}{{\sin }^{2}15}cos30}}$$$${\sin }^2\alpha =\frac{1-\cos 2\alpha }{2}\Rightarrow {\sin }^{2}15=\frac{1-\cos 30}{2}$$$$=\frac{1}{4\left(\frac{1-\cos 30}{2}\right)\cos 30}$$$$=\frac{1}{2\cos 30(1-\cos 30)}=\frac{1}{\cancel{2}\left(\frac{\sqrt{3}}{\cancel{2}}\right)(1-\frac{\sqrt{3}}{2})}$$$$=\frac{2}{2\sqrt{3}-3}\cdot \frac{2\sqrt{3}+3}{2\sqrt{3}+3}=\frac{4\sqrt{3}+6}{3}$$$$\mathcal{F}ormulas\mathcal{U}sed:\bullet \sin 2\alpha =2\sin \alpha \cos \alpha$$$$\overline{)\begin{array}{c}{\overline{)\begin{array}{c}\underset{\underset{\bullet \sin 2\alpha =2\sin \alpha \cos \alpha }{}}{\bullet {\sin }^2\alpha =\frac{1-\cos 2\alpha }{2}}\end{array}}}_{}^{}\end{array}}$$

Commented by Tawa11 last updated on 11/Oct/22

$$\mathrm{Great}\mathrm{sirs}$$

Commented by mathlove last updated on 12/Oct/22

$$thankssir$$

Answered by Rasheed.Sindhi last updated on 12/Oct/22

$$\frac{1}{\sec 15\sin 15\cos 30}=?$$$$=\frac{1}{\sec 15\sec 15\underset{―}{\cos 15\sin 15cos30}}[\because \sec 15\cos 15=1]$$$$=\frac{1}{{\sec }^{2}15\cdot \underset{―}{\frac{1}{2}\sin 30}\cos 30}[\because \cos \alpha \sin \alpha =\frac{1}{2}\sin 2\alpha ]$$$$=\frac{1}{\frac{\cancel{2}}{1+\cos 30}\cdot \frac{1}{\cancel{2}}\sin 30\cos 30}[\because {\sec }^2\alpha =\frac{2}{1+\cos 2\alpha }]$$$$=\frac{1}{\frac{1}{1+\cos 30}\cdot \frac{1}{2}\sin 60}$$$$=2(1+\cos 30)\cdot \frac{1}{\sin 60}$$$$=\{2+2\left(\frac{\sqrt{3}}{2}\right)\}.\frac{2}{\sqrt{3}}=\frac{4+2\sqrt{3}}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{6+4\sqrt{3}}{3}$$

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