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need help proving identity?
sin θ 1+cos θ
------------- +-------------- = 2cot θsecθ
1+cos θ sin θ
3 Answers
- M3Lv 71 decade agoFavorite Answer
θ has been left out for greater clarity in reading
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LHS = sin / (1+cos) + (1+cos) / sin
put LHS under a common denominator
[sin² + (1+cos)² ] / (1+cos)•sin
(sin² + 1 + cos² + 2cos) / (1+cos)•sin
2(1+cos) / (1+cos)•sin
2/sin
multiply top & bottom by cos
2cos/(sin•cos)
2cot•sec
QED
- 1 decade ago
from LHS
sin x / (1 + cos x) + (1 + cos x) / sin x
=(sin^2 x + 1 + 2 cos x + cos^2 x ) / (sin x)( 1 + cos x)
=( 2 + 2 cos x ) / (sin x)( 1 + cos x)
= 2 / sin x
= 2 cos x / (sin x cos x)
= 2cot x sec x
for more info check out http://www.a-maths-tuition.com/2010/02/proving-tri...
- Mrs.WLv 41 decade ago
we will work the LHS going for a common denominator of sinx*(1+cosx) .
[The restrictions are that the cos x does not equal -1, the cos x does not equal 0, the sinx does not equal zero]
[sin^2 (x) +1+2cosx + cos^2 (x) ]/ sinx(1+cosx)
[2 + 2cosx] / sinx* (1+cosx)
2(1 + cosx) / sinx (1+cosx)
2/ sinx
==>multiply by cosx/cosx
2cosx/sinx cosx
===> split apart into the product of 3 terms
2 * cosx/sinx * 1/ cosx
2 * cotx * sec x = 2 *cotx* secx