Prove that tan 5 theta + tan 3 theta / tan 5 theta - tan 3 theta = 4 cos 2 theta cos 4 theta
Answers
To solve this start expanding each 'tan' (tangent) term into 'sine' and 'cosine' ratio.
That is,
1) tan5x = sin5x/cos5x
2) tan3x = sin3x/cos3x
Add 1) and 2) and remember
A/B + C/D = (AD+BC)/(BD)
Using this relation we get
tan5x+tan3x = (sin5x•cos3x + cos5x•sin3x)/(cos3x•cos5x)
Doesn't the numerator on the right side of equation reminds something...
Well,
sin(a+b) = sina•cosb + sinb•cosa
Yep, put a=5 and b=3
So, we can rewrite the tangent s as,
tan5x+tan3x = sin(3x+5x)/(cos3x•cos5x)
Which is sin8x/(cos3x•cos5x)
Similarly with subtraction of 1) with 2) and using the equations
A/B-C/D = (AD-CB)/BD and
sin(a-b) = sina•cosb-sinb•cosa
You'll get
tan5x - tan3x = sin2x/(cos3x•cos5x)
Divide the two tangent terms (added tangents and subtracted tangents) on the left side of equation.
While on the right side you'll get only
sin8x/sin2x
(since cos3x•cos5x get cancelled)
Furthermore expand the numerator with,
sin2a = 2sina•cosa
This will give,
2sin4x•cos4x/sin2x
Again expand sin4x in a similar fashion,
2×2sin2x•cos2x•cos4x/sin2x
The sin2x term gets cancelled out leaving you with the required answer
(tan5x+tan3x)/(tan5x-tan3x) =
4cos2x•cos4x
Step-by-step explanation:
Let theta = x.
(tan 5x+tan 3x)/(tan 5x -tan 3x) = 4 cos 2x cos 4x.
L.H.S .
=(sin 5x /cos 5x +sin 3x /cos 3x)/( sin 5x / cos 5x - sin 3x/ cos 3x).
=[(sin 5x.cos 3x+cos 5x.sin 3x)/cos 5x.cos 3x]/[(sin 5x.cos 3x -cos 5x.sin 3x)/cos 5x.cos 3x]
=sin(5x+3x)/sin (5x-3x).
=sin8x/sin2x.
=2sin4x.cos 4x/sin 2x.
=2.2sin 2x.cos 2x.cos 4x/sin2x.
=4cos 2x.cos 4x. , Proved.