Question

please answer this question​

Answer 1

$\huge \mathfrak { \pink{solution} }\:$

Given:

$\boxed{ \bold{4 \tan( \alpha ) = 3 \:}} \\ or \ \\ \boxed{\tan( \alpha ) = \frac{3}{4} }$

To find:

$\frac{2 \sin( \alpha ) + \cos( \alpha ) }{3 \sin( \alpha ) - 2 \cos( \alpha ) }$

Divide both numerator and denominator by

$\cos( \alpha )$

we get,

$\leadsto \: \frac{2 \tan( \alpha ) + 1 }{3 \tan( \alpha ) - 2 }$

Now put the value of tan@=4/3

$\leadsto \: \frac{2 \times \frac{3}{4} + 1}{3 \times \frac{3}{4} - 2 } \\ \\ \leadsto \: \frac{ \frac{6}{4} + 1}{ \frac{9}{4} - 2} \\ \\ \leadsto \: \frac{ \frac{6 + 4}{4} }{ \frac{9 - 8}{4} } \\ \\ \leadsto \: \frac{ \frac{10}{4} }{ \frac{1}{4} } \\ \\ \leadsto \: 10$

OPTION : B

$\huge\boxed{ \red{ \bold{10}}}$

Answer 2

Given :----

• 4tanA = 3

To Find :----

$\frac{2 \sin(a) + \cos(a) }{3 \sin(a) - 2 \cos(a) }$

Formula used :-----

• sina/cosa = tana

solution :------

Dividing the Question by cosa in numerator and denominator we get, :---

$\frac{2 \tan(a) + 1 }{3 \tan(a) - 2}$

Now, putting value of tana we get,

since 4tana = 3

→ tana = 3/4

so,

$\frac{2 \times \frac{3}{4} + 1 }{3 \times \frac{3}{4} - 2} \\ \\ \frac{ \frac{3}{2} + 1}{ \frac{9}{4} - 2} \\ \\ \frac{ \frac{5}{2} }{ \frac{1}{4} } \\ \\ \frac{5}{2} \times \frac{4}{1} = 10$

Hence , value of Question is 10 (Option B)