sin (cot + tan)
= sin (cos/sin + sin/cos)
= sin(cos/sin) + sin(sin /cos)
= cos + (sin^2 / cos)
= (cos^2/cos) + (sin^2 / cos)
= (cos^2 + sin^2) / cos
= 1 / cos
= sec
2007-07-14 00:54:25
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answer #1
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answered by Mathematica 7
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sin t cot t + tan t = sec t
= sin t cos t / sin t + sin t / cos t
= cos t + sin t / cos t
= ( cos^2 t + sin t ) / cos t
its not possible to be equal
check your identities
2007-07-13 23:43:05
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answer #2
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answered by CPUcate 6
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LHS
=sin(t)(cot(t) + tan(t))
=sin(t) ( cos(t)/sin(t) +sin(t)/cos(t))
=(cos^2(t) +sin^2 (t))/cos(t)
=1/ cos(t)
=sec(t)
2007-07-13 23:19:56
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answer #3
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answered by Anonymous
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sin(t)(cot(t) + tan(t)) = sec(t)
taking left hand side
LHS
=sin(t)(cot(t) + tan(t))
=sin(t) ( cos(t)/sin(t) +sin(t)/cos(t))
=(cos^2(t) +sin^2 (t))/cos(t)
=1/ cos(t)
=sec(t)
=RHS
2007-07-14 00:04:11
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answer #4
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answered by Tubby 5
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let theta=$
sin$(cot$+tan$)
=sin$[(cos$/sin$)+(sin$/cos$)]
=sin$[(cos^2$+sin^2$)/(sin$cos$)]
=sin$(1/(sin$cos$)]
=sin$/(sin$cos$)
=1/cos$
=sec$
Hence Proved.
2007-07-13 23:54:35
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answer #5
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answered by Anonymous
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Tanx + (Cosx)/(a million+Sinx) = Secx paintings on LHS LHS = Tanx + (Cosx)/(a million+Sinx) = sinx/cosx + (Cosx)/(a million+Sinx) = [sinx(a million+Sinx) + Cos²x] / [{cosx}(a million+Sinx)] = [sinx + sin²x + cos²x] / [{cosx}(a million+Sinx)] = [sinx + a million] / [{cosx}(a million+Sinx)] = a million / {cosx} = secx = RHS
2016-10-01 14:19:42
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answer #6
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answered by Anonymous
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L.H.S
sin θ [ cos θ / sin θ + sin θ / cos θ ]
= cos θ + sin² θ / cos θ
= (cos² θ + sin² θ ) / cos θ
= 1 / cos θ
= sec θ = R.H.S.
2007-07-14 02:36:06
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answer #7
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answered by Como 7
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sinθ(cotθ+tanθ)=secθ
LHS=sinθ(cotθ+tanθ)
=sinθ(cosθ/sinθ+sinθ/cosθ)
=cosθ+sin²θ/cosθ
=(cos²θ+sin²θ)/cosθ
=1/cosθ
=secθ
=RHS
you must know (sin²θ+cos²θ)=1 and 1/cosθ=secθ
2007-07-14 01:57:07
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answer #8
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answered by Kapil 3
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sin(t)(cos(t)/sin(t)+sin(t)/cos(t))=sec(t)
sin(t)cos(t)/sin(t)+sin(t)sin(t)/cos(t)=sec(t)
cos(t)+sin^2(t)/cos(t)=sec(t)
cos^2(t)/cos(t)+sin^2(t)/cos(t)=sec(t)
cos^2(t)+ sin^2(t)=1
1/cos(t)=sec(t)
2007-07-14 01:21:14
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answer #9
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answered by epal 1
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