2sinx/sin3x +tanx/tan3x
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Answers
Given: The trigonometric term 2sinx/sin3x + tanx/tan3x
To find: The value of above expression?
Solution:
- Now we have given the term 2sinx/sin3x + tanx/tan3x
- Now we know that:
sin 3x is 3sinx - 4sin^3 x
tan 3x = 3tanx - tan^3x / 1 - 3tan^2x
- So putting these values in it, we get:
2sinx/ (3sinx - 4sin^3 x ) + tanx/ (3tanx - tan^3x / 1 - 3tan^2x)
- Simplifying it, we get:
2 / (3 - 4sin^2x) + (1 - 3tan^2x / 3 - tan^2x )
- Now converting them to sin and cos, we get:
2 / (3 - 4sin^2x) + (1 - (3 sin^2x / cos^2x) )/ (3 - (3 sin^2x / cos^2x) )
- Again converting all to sin, we get:
2 / (3 - 4sin^2x) + (1 - (3 sin^2x / 1 - sin^2x)) / (3 - (3 sin^2x / 1 - sin^2x) )
- Now simplifying, we get:
2 / (3 - 4sin^2x) + ( 1 - sin^2x - 3sin^2x / 3 - 3sin^2x - sin^2x )
2 / (3 - 4sin^2x) + (1 - 4sin^2x / 3 - 4sin^2x )
- Now denominators are same, so:
( 2 + 1 - 4sin^2x ) / 3 - 4sin^2x
3 - 4sin^2x / 3 - 4sin^2x
1
Answer:
So the value of 2sinx/sin3x + tanx/tan3x is 1.
sin3x
+
tanx
tan3x
=
2sinx
3sinx-4sin3x
+
tanx
3tanx-tan3x
1-3tan2x
=
2
3-4sin2x
+
1-3tan2x
3-tan2x
=
2
3-4sin2x
+
1-3
sin2x
cos2x
3-
sin2x
cos2x
=
2
3-4sin2x
+
1-3
sin2x
1-sin2x
3-
sin2x
1-sin2x
=
2
3-4sin2x
+
1-sin2x-3sin2x
3-3sin2x-sin2x
=
2
3-4sin2x
+
1-4sin2x
3-4sin2x
=
2+1-4sin2x
3-4sin2x
=
3-4sin2x
3-4sin2x
=1
∴
2sinx
sin3x
+
tanx
tan3x
=1.
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