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QJEEM23Maf81a: JEE Main 2023 Mathematics - Class 11 JEEM23Maf81a Solution: In a triangle ABC, if and the lengths of the sides opposite to…

In a triangle ABC, if cos⁡A+2cos⁡B+cos⁡C=2\cos \mathrm{A}+2 \cos \mathrm{B}+\cos C=2 and the lengths of the sides opposite to the angles A and C are 3 and 7 respectively, then cos⁡A−cos⁡C\mathrm{\cos A-\cos C} is equal to

Multiple Choice Options

  • A 37\frac{3}{7}
  • B 97\frac{9}{7}
  • C 107\frac{10}{7}
  • D 57\frac{5}{7}

Step-by-Step Verified Solution

amp;cos⁡A+2cos⁡B+cos⁡C=2amp;cos⁡A+cos⁡C=2(1−cos⁡B)amp;2cos⁡A+C2cos⁡(A−C2)=2×2sin⁡2B2amp;cos⁡A−C2=2sin⁡B2amp;2cos⁡B2cos⁡A−C2=4sin⁡B2cos⁡B2amp;2sin⁡(A+C2)cos⁡(A−C2)=2sin⁡Bamp;sin⁡A+sin⁡C=2sin⁡Bamp;a+c=2b(∵a=3,c=7)amp;⇒b=5\begin{aligned}& \cos A+2 \cos B+\cos C=2 \\\\& \cos A+\cos C=2(1-\cos B) \\\\& 2 \cos \frac{A+C}{2} \cos \left(\frac{A-C}{2}\right)=2 \times 2 \sin ^2 \frac{B}{2} \\\\& \cos \frac{A-C}{2}=2 \sin \frac{B}{2} \\\\& 2 \cos \frac{B}{2} \cos \frac{A-C}{2}=4 \sin \frac{B}{2} \cos \frac{B}{2} \\\\& 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)=2 \sin B \\\\& \sin A+\sin C=2 \sin B \\\\& a+c=2 b \quad(\because a=3, c=7) \\\\& \Rightarrow b=5 \quad\end{aligned}

cos⁡Aamp;−cos⁡C=b2+c2−a22bc−a2+b2−c22abamp;=25+49−970−9+25−4930amp;=6570+12=2014=107\begin{aligned}\cos A & -\cos C=\frac{b^2+c^2-a^2}{2 b c}-\frac{a^2+b^2-c^2}{2 \mathrm{ab}} \\\\& =\frac{25+49-9}{70}-\frac{9+25-49}{30} \\\\& =\frac{65}{70}+\frac{1}{2}=\frac{20}{14}=\frac{10}{7}\end{aligned}

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