If secA-tanA=3,is secA+tanA=1/3?how?
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If secA – tanA = 3
AS WE KNOW
sec² A - tan² A = 1
(sec A + tan A)(sec A -tan A)=1
(sec A + tan A)(3)=1
(sec A + tan A)=1/3
AS WE KNOW
sec² A - tan² A = 1
(sec A + tan A)(sec A -tan A)=1
(sec A + tan A)(3)=1
(sec A + tan A)=1/3
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Tysm!
Answered by
0
If secA – tanA = 3
AS WE KNOW
sec² A - tan² A = 1
(sec A + tan A)(sec A -tan A)=1
(sec A + tan A)(3)=1
(sec A + tan A)=1/3
Hence proved.
May this help u..
AS WE KNOW
sec² A - tan² A = 1
(sec A + tan A)(sec A -tan A)=1
(sec A + tan A)(3)=1
(sec A + tan A)=1/3
Hence proved.
May this help u..
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