Ριεαsε sοινε:
(1 + cotθ - cosecθ)(1 + tanθ + secθ) = 2
βεsτ αηsωεr ωiιι βε sειεςτεδ αs τhε βrαiηιiεsτ αηsωεr...
vardhmansurana1008:
ncrt question
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Answered by
6
hey mate,,,,,
here is the solution of your question
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all you need to do is expand it.
1 + tanθ + secθ+ cotθ + tanθcotθ + secθcotθ - cscθ - cscθtanθ - cscθsecθ
1 + sinθ/cosθ + 1/cosθ + cosθ/sinθ + tanθ * 1/tanθ + 1/sinθ - 1/sinθ - sinθ/cosθ - 1/sinθ * 1/cosθ
1 + 1/sinθcosθ + 1/cosθ + 1 - 1/cosθ - 1/sinθcosθ
2
I hope it will help you,,,,,,,,
here is the solution of your question
______________________________
☺️☺️☺️☺️☺️
all you need to do is expand it.
1 + tanθ + secθ+ cotθ + tanθcotθ + secθcotθ - cscθ - cscθtanθ - cscθsecθ
1 + sinθ/cosθ + 1/cosθ + cosθ/sinθ + tanθ * 1/tanθ + 1/sinθ - 1/sinθ - sinθ/cosθ - 1/sinθ * 1/cosθ
1 + 1/sinθcosθ + 1/cosθ + 1 - 1/cosθ - 1/sinθcosθ
2
I hope it will help you,,,,,,,,
Answered by
3
(1 + cotθ - cosecθ)(1 + tanθ + secθ)=2
LHS= ( 1 + cosθ/sinθ - 1/sinθ)(1 +sinθ/cosθ + 1/cosθ) -------------------
[cotx = cosx/sinx, cosecx =1/sinx, tanx=sinx/cosx, secx =cosx]
=({sinθ + cosθ -1}/sinθ)({cosθ +sinθ +1}/cosθ)
=[{(sinθ + cosθ) - 1}{(sinθ + cosθ) + 1}]/sinθcosθ
=[(sinθ + cosθ)² - 1]/sinθcosθ--------------[(a - b)(a+b)= a² +b²]
=[sin²θ + cos²θ + 2sinθcos - 1]/sinθcosθ-----------------{sin²x + cos²x =1]
= 2sinθcosθ/sinθcosθ = 2
=RHS
Hence proved.
Hope it helps.
Cheers!
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