Question

sec A- tan A=5. Then find cos A​

Answer 1

here is ur answer

sec A - tan A = 5, or

(1/cos A) - (sin A/cos A) = 5, or

1-sin A = 5 cos A, or

1-sin A = 5(1-sin^2 A)^0.5, or

Square both sides

(1-sin A)^2 = 25(1-sin^2 A)

1 - 2 sin A + sin^2 A = 25 - 25 sin^2 A, or

26sin^2 A - 2 sin A - 24 = 0, or

13 sin^2 A - sin A - 12 = 0.

sin A(1) = [+1+(1+625)^0.5]/26

= [+1+25]/26

= 26/26 = 1

Hence A(1) = 90 deg

and so sec A(1) = infinity.

sin A(2) = [+1-(1+625)^0.5]/26

= [+1-25]/26

= -24/26 = -12/13

Thus A(2) = arc sin (-12/13) = -67.38013505

cos A(2) = 0.384615384

sec A(2) = 1/0.384615384 = 2.6

Therefore, sec A = infinity or 2.6. Answer

hope it helps

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