SQRT{3}\TEXT{cosec}\,20{}^\circ -\SEC 20{}^\circ =\frac{\sqrt{3}}{\sin 20{}^\circ }-\frac{1}{\cos \,20{}^\circ }\] \[=\frac{\sqrt{3}\cos 20{}^\circ -\sin 20{}^\circ }{\sin 20{}^\circ \cos 20{}^\circ }=\frac{2\left[ \frac{\sqrt{3}}{2}\cos 20{}^\circ -\frac{1}{2}\sin \,20{}^\circ\right]}{\frac{2}{2}\sin 20{}^\circ \cos 20{}^\circ }\] \[=\frac{4\cos (20{}^\circ +30{}^\circ )}{\sin 40{}^\circ }=\frac{4\cos 50{}^\circ }{\sin 40{}^\circ }=\frac{4\sin 40{}^\circ }{\sin 40{}^\circ }=4\].

"> SQRT{3}\TEXT{cosec}\,20{}^\circ -\SEC 20{}^\circ =\frac{\sqrt{3}}{\sin 20{}^\circ }-\frac{1}{\cos \,20{}^\circ }\] \[=\frac{\sqrt{3}\cos 20{}^\circ -\sin 20{}^\circ }{\sin 20{}^\circ \cos 20{}^\circ }=\frac{2\left[ \frac{\sqrt{3}}{2}\cos 20{}^\circ -\frac{1}{2}\sin \,20{}^\circ\right]}{\frac{2}{2}\sin 20{}^\circ \cos 20{}^\circ }\] \[=\frac{4\cos (20{}^\circ +30{}^\circ )}{\sin 40{}^\circ }=\frac{4\cos 50{}^\circ }{\sin 40{}^\circ }=\frac{4\sin 40{}^\circ }{\sin 40{}^\circ }=4\].

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\[\sqrt{3}\,\text{cosec}\,{{20}^{o}}-\sec \,{{20}^{o}}=\] [IIT 1988]

Mathematics Trigonometric Identities in Mathematics . 5 months ago

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\[\SQRT{3}\TEXT{cosec}\,20{}^\circ -\SEC 20{}^\circ =\frac{\sqrt{3}}{\sin 20{}^\circ }-\frac{1}{\cos \,20{}^\circ }\] \[=\frac{\sqrt{3}\cos 20{}^\circ -\sin 20{}^\circ }{\sin 20{}^\circ \cos 20{}^\circ }=\frac{2\left[ \frac{\sqrt{3}}{2}\cos 20{}^\circ -\frac{1}{2}\sin \,20{}^\circ\right]}{\frac{2}{2}\sin 20{}^\circ \cos 20{}^\circ }\] \[=\frac{4\cos (20{}^\circ +30{}^\circ )}{\sin 40{}^\circ }=\frac{4\cos 50{}^\circ }{\sin 40{}^\circ }=\frac{4\sin 40{}^\circ }{\sin 40{}^\circ }=4\].

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Posted on 15 Aug 2024, this text provides information on Mathematics related to Trigonometric Identities in Mathematics. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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