Question

Given 3 cosA - 4 sinA = 0; evaluate without using tables: sinA + 2 cosA / 3 cosA - sin A.​

Answer 1

`\large\underline{\sf{Given\: info-}}`

3 cos A - 4 sin A = 0

To Evaluate:

`\sf(sin A + 2 cos A)/(3 cos A - sin A)`

`\large\underline{\sf{Solution-}}`

`\sf\longmapsto 3\cos A-4\sin A=0`

Transposing 4 sin A to RHS,

`\sf\longmapsto 3\cos A=4\sin A`

Transposing 4 to LHS and cos A to RHS,

`\sf\longmapsto (3)/(4)=(\sin A)/(\cos A)`

So,

`\sf\longmapsto \tan A=(3)/(4)`

Now, as
`\sf tan A = $(3)/(4)$`,

`\sf\longmapsto \sin A=(3)/(5)`

And,

`\sf\longmapsto \cos A=(4)/(5)`

We need to find that,

`\sf\longmapsto (\sin A+2\cos A)/(3\cos A-\sin A)`

So,

`\sf\longmapsto (\left((3)/(5)\right)+2\left((4)/(5)\right))/(3\left((4)/(5)\right)-\left((3)/(5)\right) )`

`\sf\longmapsto (\left((3)/(5)\right)+\left((8)/(5)\right))/(\left((12)/(5)\right)-\left((3)/(5)\right) )`

`\sf\longmapsto (\left((3+8)/(5)\right))/(\left((12-3)/(5)\right))`

So,

`\sf\longmapsto (\left((11)/(5\!\!\!/)\right))/(\left((9)/(5\!\!\!/)\right))`

So,

`\sf\longmapsto (11)/(9)`

Answer:-

`\implies\bf (sin A+2cos A)/(3cos A-sin A)=(11)/(9)`

Answer 2

11/9

Explanation:

3cosA-4sinA=0

cosA=4/3sinA

`(sinA+2cosA)/(3cosA-sinA) =(sinA+2 * (4)/(3)sin A)/(4sinA-sinA) \\=(sinA(1+(8)/(3)))/(3 sinA) \\=(3+8)/(9) \\=(11)/(9)`