sin3ϴ/sinϴ-cos3ϴ/cosϴ=2
Answers
Answered by
3
Answer:
Given: [sin(3θ)/sinθ-cos(3θ)/cos(θ)] = 2
= [sin(3θ)cos(θ)−cos(3θ)sin(θ)]/[sin(θ)cos(θ)]
= [sin(3θ−θ)]/[sin(θ)cos(θ)]
= [sin(2θ]/[sin(θ)cos(θ)]
= [2 sin(θ)cos(θ)]/[sin(θ)cos(θ)]
= 2
Answered by
0
3 is the right answer
Step-by-step explanation:
like my answer and rate it
Using trigonometric identity:
sin3θ=3sinθ−4sin
3
θ⇒sin
3
θ+sin3θ=3sinθ−3sin
3
θ
cos3θ=4cos
3
θ−3cosθ⇒cos
3
θ−cos3θ=3cosθ−3cos
3
θ
Taking L.H.S.-
cosθ
cos
3
θ−cos3θ
+
sinθ
sin
3
θ+sin3θ
=
cosθ
3cosθ−3cos
3
θ
+
sinθ
3sinθ−3sin
3
θ
=
cosθ
3cosθ(1−cos
2
θ)
+
sinθ
3sinθ(1−sin
2
θ)
=3sin
2
θ+3cos
2
θ
=3(sin
2
θ+cos
2
θ)
=3 =R.H.S
Hence Proved
Similar questions