If 3 tan (0 – 15°) = tan (0 + 15°), 0 < 0 < 90°, then prove that 0 = 45°.
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Answer:
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
∘
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