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IN AN ACUTE ANGLE TRIANGLE ABC , sec A( 1+ sec A) secB(1+secB) sec C (1+sec C ) =?

IN AN ACUTE ANGLE TRIANGLE ABC , sec A( 1+ sec A) secB(1+secB) sec C (1+sec C ) =?
 

Grade:12th pass

1 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
2 years ago
Dear student

As there is no condition consider equilateral triangle

Sec A = 2 = Sec B= Sec C

Ans is (2*3)*(2*3)*(2*3) = 216

Regards

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