tan 5 theta+ tan 3 theta/tan 5 theta- tan 3 theta=4 cos 2theta cos 4theta proveit
Answers
Answer:
hlo dear ♥️
Step-by-step explanation:
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
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Answer:
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