To prove that.... plaza answer this fastly...
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=> (cotA -cosA)/(cotA + cosA) = (cosecA-1)/. (cosecA+1)
=> (cotA -cosA)/(cotA + cosA)
=> ((cosA/sinA)-cosA) / ((cosA/sinA)-cosA)
taking cosA common from numerator nd denominator & cancelling it we get
((1/sinA)-1)/((1/sinA)+1) since 1/sinA=cosecA
therefore,
=> (cosecA-1)/(cosecA+1)
Hope it helps..
Thank You !
=> (cotA -cosA)/(cotA + cosA)
=> ((cosA/sinA)-cosA) / ((cosA/sinA)-cosA)
taking cosA common from numerator nd denominator & cancelling it we get
((1/sinA)-1)/((1/sinA)+1) since 1/sinA=cosecA
therefore,
=> (cosecA-1)/(cosecA+1)
Hope it helps..
Thank You !
THERANGER:
plzz answe my new question
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Here is the solution
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ok but plzz answer my new question
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