Question

If `6sinA + 8cosA` = 10 , how to prove that `TanA = 3/4` ?

Answer 1

See the explanation below

Explanation:

`6sinA+8cosA=10`

Dividing both sides by `10`

`3/5sinA+4/5cosA=1`

Let `cosalpha=3/5` and `sinalpha=4/5`

`cosalpha=cosalpha/sinalpha=(3/5)/(4/5)=3/4`

Therefore,

`sinAcosalpha+sinalphacosA=sin(A+alpha)=1`

So,

`A+alpha=pi/2`, `mod[2pi]`

`A=pi/2-alpha`

`tanA=tan(pi/2-alpha)=cotalpha=3/4`

`tanA=3/4`

`QED`

Answer 2

see below.

Explanation:

`or,6sinA - 10 = -8cosA`
`or, (6sinA -10)^2 = (-8cosA)^2`
`or, 36sin^2A- 2*6sinA*10 + 100 = 64cos^2A`
`or, 36sin^2A - 120sinA+100 = 64cos^2A`
`or, 36sin^2A - 120sinA + 100 = 64(1 - sin^2A)`
`or, 36sinA - 120sinA +100 = 64 - 64Sin^2A`
`or, 100 sin^2A - 120SinA + 36 = 0`
`or, (10sinA-6)^2 = 0`
`or, 10sinA - 6 =0`
`or, SinA = 6/10`
`or,SinA = 3/5 = p/h`
Using Pythagoras theorem, we get
`b^2 = h^2 - p^2`
`or, b^2 = 5^2 - 3^2`
`or, b^2 = 25 - 9`
`or, b^2 = 16`
`or, b = 4`
`so,TanA= p/b = 3/4`
Is this answer correct?

Answer 3

see solution

Explanation:

`6sinA+8cosA=10`

dividing both sides by `sqrt(6^2+8^2)`=`10`

`(6sinA)/10+8cosA/10=10/10=1`

`cosalphasinA+sinalphacosA`=1

where `tanalpha=4/3` or `alpha=53degree`

this transforms to

`sin(alpha+A)=sin90`

`alpha +A=90`

`A=90-alpha`

taking `tan`both sides

`tanA=tan(90-alpha)`

`tanA=cotalpha`
`tanA=3/4`

Answer 4

`6sinA+8cosA = 10`

`=>3sinA+4cosA = 5`

`=>(3/5)sinA + (4/5)cosA =1`

`=>(3/5)sinA + (4/5)cosA =(sinA)^2 + (cosA)^2`
`[color(red)(sin^2A + cos^2A = 1)]`

`=>(3/5) sinA + (4/5) cosA = sinA*sinA + cosA*cosA`

`=>sinA=3/5 and cosA =4/5`

Hence, `tanA = sinA/cosA =(3/5) / (4/5)=(3/5) × ( 5/4) = 3/4`