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Sec(A+B) + Sec(A-B) = 2SecA then prove that cosA = +_ underroot2cosB\2

Sec(A+B) + Sec(A-B) = 2SecA
then prove that cosA = +_ underroot2cosB\2

Grade:11

2 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
6 years ago
Dear student,
We have:
sec\left ( A+B \right )+sec\left ( A-B \right )=2sec\left ( A \right )\Rightarrow cos\left ( A-B \right )+cos\left ( A+B \right )=2sec\left ( A \right )cos\left ( A+B \right )cos\left ( A-B \right )\Rightarrow 2cos^{2}\left ( A \right )cos\left ( B \right )=-1+cos^{2}\left ( A \right )+cos^{2}\left ( B \right )\Rightarrow cos^{2}\left ( A \right )=\frac{sin^{2}\left (B \right )}{1-2cos\left ( B \right )}=4cos^{2}\left ( \frac{B}{2} \right )\left ( \frac{1-cos^{2}\left ( \frac{B}{2} \right )}{3-4cos^{2}\left ( \frac{B}{2} \right )} \right )
The next step is to just simplify and that would give the answer.
Thanks & Regards
Sumit Majumdar,
askIITians Faculty
Ph.D,IIT Delhi
Abhinav Acharya
20 Points
6 years ago
Sir, Please simplify the last step...Thank You!!

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