Question

Please solve it... Its urgent...​

Answer 1

Step-by-step explanation:

3sinbeta-4cosbeta=0

(4cotbeta-3)/(3cotbeta+4)

=(4cosbeta/sinbeta -3)/(3cosbeta/sinbeta+4)

=((4cosbeta-3sinbeta)/sinbeta)/((3cosbeta+4sinbeta)/sinbeta)

=(4cosbeta-3sinbeta)/(3cosbeta+4sinbeta)

=-(3sinbeta-4cosbeta)/(3cosbeta+4sinbeta)

=0/(3cosbeta+4sinbeta)

=0

Answer 2

given that :-

$\sf \bf \: \: 3 \sin\beta - 4 \cos \beta = 0$

to find :-

$\sf \: the \:value \: of \: \: \: \: \dfrac{4 \cot \beta - 3 }{3 \cot \beta + 4}$

Solution :-

$\sf \: 3 \sin \beta - 4 \cos \beta = 0 \\ \\ \sf\: 3 \sin \beta = 4 \cos \beta \\ \\ \sf \red {or} \:\:\: \:\:\:\dfrac{ \sin\beta }{ \cos \beta } = \dfrac{4}{3} \\ \\ \sf \red {or} \:\:\: \:\:\: \tan \beta = \dfrac{4}{3} \\ \\ \sf \: \red {or} \:\:\:\:\:\:\frac{1}{ \cot \beta } = \dfrac{4}{3} \\ \\ \sf \red {or} \:\:\: \:\:\:\cot \beta = \dfrac{3}{4}$

now put the value of cot beta = 3/4 in

$\sf \dfrac{4 \cot \beta - 3 }{3 \cot \beta + 4}$

we get :-

$\sf \red \implies \: \dfrac{4 \cot \beta - 3 }{3 \cot \beta + 4} \\ \\ \sf \red\implies \dfrac{4 \times \frac{3}{4} - 3}{3 \times \frac{3}{4} + 4} \\ \\ \sf \red \implies \: \dfrac{ \cancel4 \times \frac{3}{ \cancel4} - 3}{3 \times \frac{3}{4} + 4} \\ \\ \sf \red\implies \: \dfrac{3 - 3}{ \frac{9 + 16}{4} } \\ \\ \sf \red\implies \frac{0 \times 4}{25} \\ \\ \small{\blue{\fcolorbox{blue}{lime}{\boxed{\orange{\bf{\fcolorbox{blue}{black} { { \blue{ 0 }}}}}}}}}$

hence value of $\sf \dfrac{4 \cot \beta - 3 }{3 \cot \beta + 4}$ is 0