If 6sinA + 8cosA6sinA+8cosA = 10 , how to prove that TanA = 3/4tanA=34 ?

4 Answers
Mar 29, 2018

See the explanation below

Explanation:

6sinA+8cosA=106sinA+8cosA=10

Dividing both sides by 1010

3/5sinA+4/5cosA=135sinA+45cosA=1

Let cosalpha=3/5cosα=35 and sinalpha=4/5sinα=45

cosalpha=cosalpha/sinalpha=(3/5)/(4/5)=3/4cosα=cosαsinα=3545=34

Therefore,

sinAcosalpha+sinalphacosA=sin(A+alpha)=1sinAcosα+sinαcosA=sin(A+α)=1

So,

A+alpha=pi/2A+α=π2, mod[2pi]

A=pi/2-alpha

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

tanA=3/4

QED

Mar 29, 2018

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?

Mar 29, 2018

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 tanboth sides

tanA=tan(90-alpha)

tanA=cotalpha
tanA=3/4

Mar 29, 2018

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