Question

If 3 tan theta = 4 find the value of 5 sin theta - 3 cos theta/5 sin theta + 2 cos theta

Answer 1

solution:

5tan\theta=4

tan\theta=4/5

=perpandicular/base

therefore,

hypotenuse=\sqrt{base^2+perpendicular^2}

=\sqrt{5^2+4^2}

=\sqrt{25+16}=\sqrt{41}

sin\theta=4/\sqrt{41}

cos\theta=5/\sqrt{41}

then,

\frac{5sin\theta-3cos\theta}{5sin\theta+2cos\theta}=\frac{5*4/\sqrt{41}-3*5/\sqrt{41}}

{5*4/\sqrt{41}+2*5/\sqrt{41}

}

=\frac{5/\sqrt{41}}{30/\sqrt{41}}

=5/30=1/6

Answer 2

$\boxed{ \underline {\sf {\green{Given :-}}}} \\ \\ \bold{5 \tan \theta \: = 4 } \\ \implies{ \tan \theta \: = \: \frac{4}{5} \: \: \: \: \: \: \: \: \: \: \: ....(i) } \\ \\ \boxed{ \underline {\sf {\green{solution :-}}}} \\ \bold{Now,} \\ \\ \sf \implies{ \frac{5 \sin \theta - 3 \cos \theta }{5 \sin \theta + 2 \cos \theta } = \frac{5 \frac{ \sin \theta }{ \cos\theta } - 3 \frac{ \cos \theta}{ \cos\theta} }{5 \frac{ \sin\theta}{ \cos\theta } + 2 \frac{ \cos\theta}{ \cos\theta} } } \\ \\ \: \: \: \: \: \: \: \: \: \: \sf[Divide \: the \: numerator \: and \: denominator \: by \: cos \theta] \\ \\ \sf \implies{ \frac{5 \tan \theta - 3 }{5 \tan \theta + 2 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (∵ \: \tan \theta \: = \frac{ \sin \theta }{ \cos \theta }) } \\ \\ \sf \implies{ \frac{5 . \frac{4}{5} - 3 }{5. \frac{4}{5} + 2 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [ \: using \: (i) \: ]} \\ \\ \sf \purple{ \implies{\frac{4 - 3}{4 + 2} = \frac{1}{6} .}}$