Question

If 3cotθ=1, then find the value of

(3cosθ-4sinθ)/(5sinθ+4cosθ).​

Answer 1

Required Answer:-

Given:

• 3cot(x) = 1

To find:

• Value of (3cos(x) - 4sin(x))/(5sin(x) + 4cos(x))

Solution:

Given that,

$\rm 3 \cot(x) = 1$

$\rm \implies \cot(x) = \dfrac{1}{3}$

Now,

$\rm \dfrac{3 \cos(x) - 4 \sin(x) }{5 \sin(x) + 4 \cos(x) }$

Dividing the numerator and denominator by sin(x), we get,

$\rm = \dfrac{ \dfrac{1}{ \sin(x) } \bigg \{ 3 \cos(x) - 4 \sin(x) \bigg\}}{ \dfrac{1}{ \sin(x) } \bigg \{5 \sin(x) + 4 \cos(x) \bigg\} }$

$\rm = \dfrac{ \bigg \{ 3 \dfrac{ \cos(x) }{ \sin(x) } - 4 \dfrac{\sin(x) }{ \sin(x) } \bigg\}}{ \bigg \{5 \dfrac{\sin(x) }{ \sin(x) }+ 4 \dfrac{\cos(x)}{ \sin(x) } \bigg\} }$

$\rm = \dfrac{ 3 \cot(x) - 4 }{5 + 4 \cot(x) }$

$\rm = \dfrac{1 - 4 }{5 + 4 \times \dfrac{1}{3} }$

$\rm = \dfrac{ - 3 }{5 + \dfrac{4}{3} }$

$\rm = \dfrac{ - 3 }{ \dfrac{15 + 4}{3} }$

$\rm = \dfrac{ - 3 }{ \dfrac{19}{3} }$

$\rm = \dfrac{ -3 \times 3}{19}$

$\rm = \dfrac{ -9}{19}$

Hence,

$\rm \mapsto \dfrac{3 \cos(x) - 4 \sin(x) }{5 \sin(x) + 4 \cos(x) } = \dfrac{ - 9}{19}$

Note:

• sin(x)/cos(x) = tan(x)
• cos(x)/sin(x) = 1/tan(x) = cot(x)

Answer 2

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