Question

If 4 tanβ = 3, then 4 −3 cos 4 sin +3 cos ​

Answer 1

SOLUTION

GIVEN

4tanβ = 3

TO DETERMINE

$\displaystyle \sf{ \frac{4 \sin \beta - 3 \cos \beta }{4 \sin \beta + 3 \cos \beta } }$

EVALUATION

Here it is given that

4tanβ = 3

Now

$\displaystyle \sf{ \frac{4 \sin \beta - 3 \cos \beta }{4 \sin \beta + 3 \cos \beta } }$

Dividing both of the numerator and denominator by cosβ we get

$\displaystyle \sf{ = \frac{4 \dfrac{\sin \beta}{ \cos \beta} - 3 \dfrac{ \cos \beta}{ \cos \beta} }{4 \dfrac{\sin \beta}{ \cos \beta} + 3 \dfrac{ \cos \beta}{ \cos \beta}} }$

$\displaystyle \sf{ = \frac{4 \tan \beta - 3}{4 \tan \beta + 3} }$

$\displaystyle \sf{ = \frac{3 - 3}{3 + 3} }$

$\displaystyle \sf{ = \frac{0}{6} }$

$\displaystyle \sf{ = 0 }$

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