In `triangleABC, (cos2A)/(a^(2))-(cos2B)/(b^(2))=`A. `b^(2)-a^(2)`B. ` – InterviewSolution

InterviewSolution

1.	In `triangleABC, (cos2A)/(a^(2))-(cos2B)/(b^(2))=`A. `b^(2)-a^(2)`B. `a^(2)-b^(2)`C. `(1)/(b^(2))-(1)/(a^(2))`D. `(1)/(a^(2))-(1)/(b^(2))`
Answer» Correct Answer - D

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