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( * ) If (secA + tan A) (sec B+ tanB) (secC + tan C) = (sec A – tan A) (sec B – tan B) (sec C – tan C) = k then k =

( * )
If (secA + tan A) (sec B+ tanB) (secC + tan C) = (sec A – tan A) (sec B – tan B) (sec C – tan C) = k then k =

Grade:11

1 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
9 years ago
Hello student, Please find the answer to your question below

(secA + tan A) (sec B+ tanB) (secC + tan C) = (sec A – tan A) (sec B – tan B) (sec C – tan C)
= (1+sinA/ cosA)(1+sinB/ cosB)(1+sinC/ cosC) = (1-sinA/ cosA)(1-sinB/ cosB)(1-sinC/ cosC) = k

So (1+sinA)(1+sinB)(1+sinC) = (1-sinA)(1-sinB)(1-sinC)
{(1+sinA)(1+sinB)(1+sinC)}^2 = (1-sinA)(1-sinB)(1-sinC)(1+sinA)(1+sinB)(1+sinC)
{(1+sinA)(1+sinB)(1+sinC)}^2 = cos^2 A cos^2 B cos^2 C
taking root both side
{(1+sinA)(1+sinB)(1+sinC)} = cos A cos B cos C

So
{(1+sinA)(1+sinB)(1+sinC)} /cos A cos B cos C = 1 = k

So k =1

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