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secA=13/5 ,show that 2sinA-3cosA/4sinA-9cosA=3

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secA=13/5 ,show that 2sinA-3cosA/4sinA-9cosA=3

Posted by Shivani Singh 7 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Akhilesh Gupta 7 years, 3 months ago

Given that SecA=13/5. Then we know that Sec=hypotenius/Base, I.e,13/5 Then we find perpendicular =12 Thus, 2 sinA - 3cosA /4sinA -9cosA =3 {(2 ✖ 12/13 - 3 ✖ 5/13)/(4 ✖ 12/13 - 9 ✖ 5/13)} = 3 or. {24 -15/48-45}=3 or. { 9/3}= 3 Or. 3=3 Hence LHS=RHS

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