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( m+2 )sinA + (2m – 1)cosA = ( 2m+1 ) then find tanA. (the answer is 3/4)

( m+2 )sinA + (2m – 1)cosA
= ( 2m+1 )
then find tanA.    
(the answer is 3/4)

Grade:11

3 Answers

JAGADEESH
36 Points
9 years ago
(m+2) Sin A +(2m-1) Cos A = 2m+1--------1
we know that ,
 Sin A =(2 tan(A/2))/(1+tan2(A/2) and Cos A = (1-tan2(A/2))/(1+tan2(A/2)
let tan(A/2) = k and it in first equation
We get,
  (m+2)(2k)/(1+k2)  + (2m-1)(1-k2)/(1+k2) = 2m+1
(2m+4)k+(2m-1)-(2m-1)k2 = (2m+1)(1+k2)
(4m)k2-(2m+4)k+2=0  ---------2
By solving eq 2 we get,
k=1/m or k=0.5
tan(A/2)=1/m or tan(A/2) =0.5
A/2 = tan ⁻¹(1/m) 0r A/2 =tan ⁻¹(0.5)
A=2tan ⁻¹(1/m) or 53.130102354
 
tan A=tan (2tan ⁻¹(1/m) or 4/3
 
Kaustubh Nayyar
27 Points
9 years ago
thank you
 
Rishabh
11 Points
6 years ago
(m+2)sin A + (2m−1) cos A = (2m+1)⇒(m+2)sin Acos A + (2m−1) cos Acos A = (2m+1)×1cos A⇒(m+2)tan A + (2m−1) = (2m+1) sec A⇒[(m+2)tan A + (2m−1)]2 = [ (2m+1) sec A]2⇒(m+2)2 tan2A + (2m−1)2 + 2(2m−1)(m+2) tan A = (2m+1)2 . sec2A⇒(m+2)2 tan2A + (2m−1)2 + 2(2m−1)(m+2) tan A = (2m+1)2 (1+tan2A)⇒[(m+2)2 + (2m+1)2]tan2A + 2(2m−1)(m+2) tan A + (2m−1)2 − (2m+1)2 = 0⇒3(1−m2) tan2A + (4m2+6m−4) tan A − 8m = 0⇒(3tan A − 4)[(1−m2)tan A + 2m] = 0⇒(3tan A − 4) = 0 or [(1−m2)tan A + 2m] =0⇒tan A = 43 or tan A = 2mm2−1

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