If 3 × tan(x – 15) = tan(x + 15), then the value of x is
Answers
Answer:
Let t=theta-15 degrees. Further in this text angles are in degrees.
Then the equation becomes 3tan(t) = tan(t+30)
3tan(t) = (tan(t) +tan(30))/(1-tan(t)tan(30))
3tan(t) = (tan(t) +sqrt(3)/3)/(1-tan(t)sqrt(3)/3))
3tan(t) = (3tan(t) + sqrt(3))/ (3tan(t) - sqrt(3)
Let 3 tan(t) = u
u=(u+sqrt(3))/(u-sqrt(3))
u^2-usqrt(3) = u-swrt(3)
u^2 - (1+sqrt(3))u +swrt(3) =0
Solving this quadratic equation we have u1 = 1, u2=sqrt(3).
Thus, we have tan(t) = 1/3 and tan(t) = sqrt(3)/3.
So, one series of solutions is t= arctan(1/3) + 180*n,
the other one is t= 30+180*k. k and n are arbitrary integers.
Finally, theta = arctan(1/3) + 15+180*n,
theta = 45+180*k, where n and k are arbitrary integers.
It is important to note that arctan(1/3)+15 , nor arctan(1/3) -15 is equal to 90 degrees, so both left hand side and right hand side of the equation are defined.
Solution -
3tan(θ−15
∘
)=tan(θ+15
∘
)
3=
tan(θ−15
∘
)
tan(θ+15
∘
)
3−1
3+1
=
tan(θ+15
∘
−tan(θ−15
∘
)
tan(θ+15
∘
)+tan(θ−15
∘
)
2=
cos(θ+15
∘
)
sin(θ+15
∘
)
−
cos(θ−15
∘
)
sin(θ−15
∘
)
cos(θ+15
∘
)
sin(θ+15
∘
)
+
cos(θ−15
∘
)
sin(θ−15
∘
)
2=
sin(θ+15
∘
)cos(θ−15
∘
)−sin(θ−15
∘
)cos(θ+15
∘
)
sin(θ+15
∘
)cos(θ−15
∘
)+sin(θ−15
∘
)cos(θ+15
∘
)
Apply sin(A+B)=sinAcosB+cosAsinB
sin(A−B)=sinAcosB−cosAsinB
2=
sin(30
∘
)
sin(2θ)
sin2θ=1
2θ=2nπ+
2
π
θ=nπ+
4
π
in 0<θ<90
∘
−θ=45
∘
