Question

If 4Sin A - 3 Cos A=0 find the valu of 3 Sin A - 4 Cos A

Answer 1

${\huge{\red{\sf{Given}}}}\begin{cases}\leadsto \bf{4sinA -3cosA=0}\\\leadsto\bf{3sinA-4cosA=? }\end{cases}$

${\huge{\red{\sf{To\:Find}}}}\begin{cases}\leadsto \bf{Value\:of\:3sinA-4cosA}\end{cases}$

$\huge\red{\underline{\bf{\green{Answer}}}}$

$\sf{\red{\hookrightarrow Taking\: the\:first\:given\:equ^{n}}}$

$\sf{\implies 4sinA - 3 cosA=0}$

$\sf{\implies 4sinA=3cosA}$

$\sf{\implies \dfrac{sinA}{cosA}=\dfrac{3}{4}}$

$\sf{\purple{\longmapsto tanA=\dfrac{3}{4}}}$

$\rule{200}1$

$\bf{\pink{\hookrightarrow For\: figure\:refer\:to\: attachment.}}$

$\sf{\pink{In \:right\:\triangle ABC}}$

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

$\sf{\mapsto AC^{2}=(3x)^{2}+(4x)^{2}}$

$\sf{\mapsto AC^{2}=9x^{2}+16x^{2}}$

$\sf{\mapsto AC^{2}=25x^{2}}$

$\sf{\pink{\leadsto AC=5x}}$

$\rule{200}3$

$\sf{\purple{sinA=\dfrac{perpendicular}{hypotenuse}=\dfrac{3x}{5x}=\frac{3}{5}}}$

$\sf{\blue{cosA=\dfrac{base}{hypotenuse}=\dfrac{4x}{5x}=\frac{4}{5}}}$

${\underline{\sf{\red{Now,}}}}$

$\bf{\green{\leadsto 3sinA-4cosA}}$

$\sf{\implies 3sinA-4cosA=3\times\dfrac{3}{5}-4\times\dfrac{4}{5}}$

$\sf{\implies 3sinA-4cosA=\dfrac{9}{5}-\dfrac{16}{5}}$

$\sf{\implies 3sinA-4cosA =\dfrac{-16+9}{5}}$

${\underline{\orange{\boxed{\bf{\longmapsto 3sinA-4cosA=\dfrac{-7}{5}}}}}}$

$\underline{\sf{\pink{\therefore The\: required\: answer\:is\:3sinA-4cosA=\dfrac{-7}{5}}}}$

Answer 2

hope it helps .............