Question

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If 4 tan$\beta$= 3, then$\frac{4 \sin\beta - 3 \cos \beta }{4 \sin\beta + 3 \cos \beta}$ is equal to?​

Answer 1

Given :-

→ 4 tan β = 3

→ $\tan \beta = \frac{3}{4}$

Solution :-

$→ \tan \beta = \frac{3}{4} = \frac{b}{p} = \frac{AB}{BC}$

• AB = 3 cm
• BC = 4 cm
• AC = ?

By Pythagoras Theorem,

$\sf{→{h}^{2} = {p}^{2} + {b}^{2}}$

$\sf{→ {AC}^{2} = {AB}^{2} + {BC}^{2}}$

$\sf{→ {AC}^{2} = {4}^{2} + {3}^{2}}$

$\sf{→ {AC}^{2} = 16 + 5}$

$\sf{→ {AC}^{2} = 25}$

$\sf{→ AC = \sqrt{25}} = 5 \: cm$

Equation :-

$→ \boxed{\dfrac{4 \sin\beta - 3 \cos \beta }{4 \sin\beta + 3 \cos \beta}}$

$\sf{→ \frac{4 \times \frac{AB}{AC} - 3 \times \frac{BC}{AC}}{4 \times \frac{AB}{AC} + 3 \times \frac{BC}{AC}}}$

$→ \frac{4 \times \frac{4}{5} - 3 \times \frac{3}{5} }{4 \times \frac{4}{5} - 3 \times \frac{3}{5}}$

$→ \frac{ \frac{16}{5} - \frac{9}{5}}{ \frac{16}{5} + \frac{9}{5} }$

$→ \frac{ \frac{7}{ \cancel{5}}}{ \frac{25}{ \cancel{5}}}$

$→ \boxed {\dfrac{7}{25}}$

Hence, $\frac{4 \sin\beta - 3 \cos \beta }{4 \sin\beta + 3 \cos \beta}$ is equal to $\frac{7}{25}$.

Answer 2

Step-by-step explanation:

Given :-

→ 4 tan β = 3

→ $\tan \beta = \frac{3}{4}$

Solution :-

$→ \tan \beta = \frac{3}{4} = \frac{b}{p} = \frac{AB}{BC}$

AB = 3 cm

BC = 4 cm

AC = ?

By Pythagoras Theorem,

$\sf{→{h}^{2} = {p}^{2} + {b}^{2}}$

$\sf{→ {AC}^{2} = {AB}^{2} + {BC}^{2}}$

$\sf{→ {AC}^{2} = {4}^{2} + {3}^{2}}$

$\sf{→ {AC}^{2} = 16 + 5}$

$\sf{→ {AC}^{2} = 25}$

$\sf{→ AC = \sqrt{25}} = 5 \: cm$

Equation :-

$→ \boxed{\dfrac{4 \sin\beta - 3 \cos \beta }{4 \sin\beta + 3 \cos \beta}}$

$\sf{→ \frac{4 \times \frac{AB}{AC} - 3 \times \frac{BC}{AC}}{4 \times \frac{AB}{AC} + 3 \times \frac{BC}{AC}}}$

$→ \frac{4 \times \frac{4}{5} - 3 \times \frac{3}{5} }{4 \times \frac{4}{5} - 3 \times \frac{3}{5}}$

$→ \frac{ \frac{16}{5} - \frac{9}{5}}{ \frac{16}{5} + \frac{9}{5} }$

$→ \frac{ \frac{7}{ \cancel{5}}}{ \frac{25}{ \cancel{5}}}$

$→ \boxed {\dfrac{7}{25}}$

Hence, $\frac{4 \sin\beta - 3 \cos \beta }{4 \sin\beta + 3 \cos \beta}$ is equal to $\frac{7}{25}$.