Question

If tanA +secA = 3/2 then find sinA,tanA and secA

Answer 1

tan A +sec A = 3/2
sin A/cos A + 1/cos A = 3/2
(1 + sin A)/cos A = 3/2

Squaring,
(1+ sin A )^2/cos^2 A = 9/4

(1+ sin A )^2/(1-sin^2 A)
= 9/4

(1+sin A )^2 / [(1-sin A)(1+sin A)] = 9/4

(1+sin A)/(1-sin A) = 9/4

4 + 4sin A = 9 - 9sin A
13sin A = 5
sin A = 5/13

cos A = √(1 - sin^2 A)
= √(1 - 25/169)
= √(169 - 25)/169
= 12/13

tan A = sin A / cos A
= (5/13) / (12/13)
= 5/12

sec A = 1/cos A
= 13/12