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

4 Answers
Mar 29, 2018

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#

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 #tan#both 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#