If 4cosA-5sinA=0 then find the value of 3sinA -2cosA/5sinA+3cosA
Answers
Answer:
= ?
solution:
Squaring and adding equations (1) and (2), we get
(3\cos A - 4\sin A)^2 + (4\cos A + 3\sin A)^2(3cosA−4sinA)
2
+(4cosA+3sinA)
2
= 5^25
2
+ x^{2}x
2
⇒ 25 + x^{2}x
2
= (3\cos A)^2 + (4\sin A)^2 + 2(3\cos A)(4\sin A)+ (4\cos A)^2 + (3\sin A)^2-2(3\cos A)(4\sin A)(3cosA)
+(4sinA)
2
+2(3cosA)(4sinA)+(4cosA) 2
+(3sinA)
2 −2(3cosA)(4sinA)
⇒ 25 + x^{2}x
2 = 9\cos^2 A +16\sin^2 A + 16\cos^2+ 9\sin^2 A9cos
2
A+16sin
2
A+16cos
2 +9sin 2 A ⇒ 25 + x^{2}x 2
= 9(\cos^2 A+\sin^2 A) +16(\cos^2 A+\sin^2 A)9(cos 2
A+sin
2
A)+16(cos
2
A+sin
2
A)
trigonometric identity:
\cos^2 Acos
2
A + \sin^2 Asin
2
A = 1
25 + x^{2}x
2
= 9(1) + 16(1)
⇒ 25 + x^{2}x
2 = 9 + 16
⇒ x^{2}x
2
= 25 - 25
⇒ x^{2}x
2
= 0
⇒ x = 0
∴ 4\cos AcosA + 3\sin AsinA = 0
Thus, if 3\cos AcosA - 4\sin AsinA = 5, then the value of "4\cos AcosA + 3\sin AsinA = 0".
Answer:
4cosa= 5sina
4/5= tana
√4^2+5^2= √41 =h
2/√41/35/√41= 2/35 is correct answer I hope you are like my answer,