if secA-tanA= 5/2 then secA+tanA=?
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2
sec^2A - Tan^2A = 1
(SecA + TanA ) ×(SecA - TanA) =1
SecA + TanA = 1/SecA - TanA
SecA + TanA = 1/5/2 = 2/5
hope it helped you...
(SecA + TanA ) ×(SecA - TanA) =1
SecA + TanA = 1/SecA - TanA
SecA + TanA = 1/5/2 = 2/5
hope it helped you...
Answered by
1
HEY.......!!! here is ur answer........☺️☺️☺️
Given that, secA–tanA = 5/2......(1)
Let, secA+tanA = x.......(2)
Now on multiplying equation (1) to (2)...
=> (secA–tanA)(secA+tanA) = 5/2.x
=> sec²A–tan²A = 5x/2
{ Because a²–b²=(a–b)(a+b) }
=> 1 = 5x/2
{ Because sec²A = 1+tan²A => sec²A–tan²A = 1 }
=> x = 2/5
=> secA+tanA = 2/5 { from equation (2) }
I hope it will help you.......✌️✌️✌️
Given that, secA–tanA = 5/2......(1)
Let, secA+tanA = x.......(2)
Now on multiplying equation (1) to (2)...
=> (secA–tanA)(secA+tanA) = 5/2.x
=> sec²A–tan²A = 5x/2
{ Because a²–b²=(a–b)(a+b) }
=> 1 = 5x/2
{ Because sec²A = 1+tan²A => sec²A–tan²A = 1 }
=> x = 2/5
=> secA+tanA = 2/5 { from equation (2) }
I hope it will help you.......✌️✌️✌️
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