COS C\cos (A-B)}{1+\cos (A-C)\cos B}=\frac{1-\cos (A+B)\cos (A-B)}{1-\cos (A-C)\cos (A+C)}\] Þ \[\frac{1-{{\cos }^{2}}A+{{\SIN }^{2}}B}{1-{{\cos }^{2}}A+{{\sin }^{2}}C}=\frac{{{\sin }^{2}}A+{{\sin }^{2}}B}{{{\sin }^{2}}A+{{\sin }^{2}}C}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}+{{c}^{2}}}\].

"> COS C\cos (A-B)}{1+\cos (A-C)\cos B}=\frac{1-\cos (A+B)\cos (A-B)}{1-\cos (A-C)\cos (A+C)}\] Þ \[\frac{1-{{\cos }^{2}}A+{{\SIN }^{2}}B}{1-{{\cos }^{2}}A+{{\sin }^{2}}C}=\frac{{{\sin }^{2}}A+{{\sin }^{2}}B}{{{\sin }^{2}}A+{{\sin }^{2}}C}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}+{{c}^{2}}}\].

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In triangle \[ABC,\]\[\frac{1+\cos (A-B)\cos C}{1+\cos (A-C)\cos B}=\]

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

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\[\frac{1+\COS C\cos (A-B)}{1+\cos (A-C)\cos B}=\frac{1-\cos (A+B)\cos (A-B)}{1-\cos (A-C)\cos (A+C)}\] Þ \[\frac{1-{{\cos }^{2}}A+{{\SIN }^{2}}B}{1-{{\cos }^{2}}A+{{\sin }^{2}}C}=\frac{{{\sin }^{2}}A+{{\sin }^{2}}B}{{{\sin }^{2}}A+{{\sin }^{2}}C}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}+{{c}^{2}}}\].

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Posted on 15 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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