Question

if 3sinA+4cosA=5 then prove tanA=3/4

Answer 1

Answer:

3 sinA+4cosA=5

3sinA+1/4sin A=5

(12sinA²+1)

Answer 2

AnswEr :

$\bold{Condition} \begin{cases} \sf{Given : 3 \sin(A) + 4 \cos(A)} \\ \sf{To \: Prove : \tan(A)= \dfrac{3}{4} }\end{cases}$

• Proof :

$\longrightarrow \rm 3 \sin(A) + 4 \cos(A) = 5$

⠀⠀⠀⠀⋆ Dividing Each term by cos (A)

$\longrightarrow \rm \dfrac{3 \sin(A)}{\cos(A)} + \dfrac{4 \cancel{\cos(A)}}{ \cancel{\cos(A)}} = \dfrac{5}{\cos(A)}$

⠀⠀⠀⠀⋆ sin (A) / cos (A) = tan (A)

⠀⠀⠀⠀⋆ 1 / cos (A) = sec (A)

$\longrightarrow \rm 3 \tan(A) + 4 = 5 \sec(A)$

⠀⠀⠀⠀⋆ squaring both sides

$\longrightarrow \rm (3 \tan(A) + 4) ^{2} = (5 \sec(A) )^{2}$

⠀⠀⠀⠀⋆ (a + b)² = a² + b² + 2ab

$\longrightarrow \small\rm (3 \tan(A))^{2} + (4) ^{2} + 2 \times 3 \tan(A) \times 4= (5 \sec(A) )^{2}$

$\longrightarrow \rm 9 \tan^{2}(A) + 16 + 24\tan(A)= 25 \sec^{2}(A)$

⠀⠀⠀⠀⋆ sec² (A) = 1 + tan² (A)

$\longrightarrow \rm 9 \tan^{2}(A) + 16 + 24\tan(A)= 25 (1 + \tan^{2}(A))$

$\longrightarrow \rm 9 \tan^{2}(A) + 16 + 24\tan(A)= 25+ 25\tan^{2}(A)$

$\longrightarrow \rm 0= 25+ 25\tan^{2}(A) - 9 \tan^{2}(A) - 16 - 24\tan(A)$

$\longrightarrow \rm 16\tan^{2}(A) - 24\tan(A) + 9 = 0$

$\longrightarrow \rm (4\tan(A))^{2} - 2 \times 4\tan(A) \times 3- {(3)}^{2} = 0$

⠀⠀⠀⠀⋆ a² + b² - 2ab = (a - b)²

$\longrightarrow \rm (4\tan(A) - 3)^{2} = 0$

$\longrightarrow \rm 4\tan(A) - 3 = 0$

$\longrightarrow \rm 4\tan(A) = 3$

$\longrightarrow \rm \boxed{\tan(A) = \dfrac{3}{4} }$