if 3cosA-4SinA=0; evaluate sinA+2cosA/3cosA-sinA
Answers
Answer:
Step-by-step explanation:
3cosA-4sinA=0
3cosA=4sinA
cosA=4sinA/3
by multiplying with conjugate
[(sinA+2cosA)(3cosA+sinA)]/[9cos²A-sin²A]
=> [3sinAcosA+sin²A+6cos²A+2sinAcosA]/[9cos²A-sin²A]
=>[sinA(4sinA)+sin²A+2(4sinA)²+2sinAcosA]/[3(4sin²A-sin²A)]
=>[4sin²A+sin²A+32sin²A+2sinAcosA]/[3sin²A]
=>[37sin²A+2sinAcosA]/3sin²A
=>37sin²A+2sinA(4sinA/3)/3sin²A
=>[(117sin²A+8sin²A)/3]/3sin²A
=>[125sin²A]/9sin²A = 125/9
the answer is 125/9
hope this is helpful to you
if you want more problems to be solve follow me or put your questions on the comments
Answer:
3cosA-4sinA=0. 3cosA=4SinA. 3/4=sinAcosA..tanA=3/4...
sinA+2cosA/3cosA.sinA=sinA+2cosA/3sinA...
Step-by-step explanation:
sonA/3sinA +2cosA/3sinA=1/3+2/3cotA....=1/3+2/3×1/tanA=1/3+(2/3×4/3)=1/3+8/9=33/27=11/9.