C=2B\Rightarrow B={{60}^{o}}\],  \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] SINCE \[B={{60}^{o}}\]\[\Rightarrow AC={{a}^{2}}+{{c}^{2}}-{{b}^{2}}\]             Þ \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-ac\] Therefore \[\frac{a+c}{\sqrt{{{a}^{2}}-ac+{{c}^{2}}}}=\frac{a+c}{b}=\frac{\sin A+\sin C}{\sin B}\] \[=\frac{2\sin \frac{A+C}{2}\cos \frac{A-C}{2}}{2\sin \frac{B}{2}\sin \frac{A+C}{2}}=\frac{\cos \frac{A-C}{2}}{\sin \frac{B}{2}}\] \[=\frac{\cos \frac{A-C}{2}}{\sin {{30}^{o}}}\Rightarrow 2\cos \frac{A-C}{2}\].

"> C=2B\Rightarrow B={{60}^{o}}\],  \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] SINCE \[B={{60}^{o}}\]\[\Rightarrow AC={{a}^{2}}+{{c}^{2}}-{{b}^{2}}\]             Þ \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-ac\] Therefore \[\frac{a+c}{\sqrt{{{a}^{2}}-ac+{{c}^{2}}}}=\frac{a+c}{b}=\frac{\sin A+\sin C}{\sin B}\] \[=\frac{2\sin \frac{A+C}{2}\cos \frac{A-C}{2}}{2\sin \frac{B}{2}\sin \frac{A+C}{2}}=\frac{\cos \frac{A-C}{2}}{\sin \frac{B}{2}}\] \[=\frac{\cos \frac{A-C}{2}}{\sin {{30}^{o}}}\Rightarrow 2\cos \frac{A-C}{2}\].

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In triangle\[ABC\]if \[A+C=2B\], then \[\frac{a+c}{\sqrt{{{a}^{2}}-ac+{{c}^{2}}}}\]is equal to [UPSEAT 1999]

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 10 months ago

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\[A+C=2B\Rightarrow B={{60}^{o}}\],  \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] SINCE \[B={{60}^{o}}\]\[\Rightarrow AC={{a}^{2}}+{{c}^{2}}-{{b}^{2}}\]             Þ \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-ac\] Therefore \[\frac{a+c}{\sqrt{{{a}^{2}}-ac+{{c}^{2}}}}=\frac{a+c}{b}=\frac{\sin A+\sin C}{\sin B}\] \[=\frac{2\sin \frac{A+C}{2}\cos \frac{A-C}{2}}{2\sin \frac{B}{2}\sin \frac{A+C}{2}}=\frac{\cos \frac{A-C}{2}}{\sin \frac{B}{2}}\] \[=\frac{\cos \frac{A-C}{2}}{\sin {{30}^{o}}}\Rightarrow 2\cos \frac{A-C}{2}\].

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Posted on 17 Sep 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). 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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