cotA +tanA=cosecAsecA
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Answered by
2
hii...
cotA+tanA=L.H.S
cosA/sinA+sinA/cosA
sin^2A+cos^A/sinAcosA
1/sinAcosA
cosecAsecA
=R.H.S
HENCE PROVED...
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Answered by
0
Step-by-step explanation:
I.e; tanA=(sinA/cosA) and cotA=(1/tanA)
.'.tanA+cotA=(sinA/cosA)+(1/(sinA/cosA)
.'.(sinA/cosA)+(cosA/sinA)
L.C.M
sin^2A+cos^2A/(cosA.sinA)
but. (sin^2A+cos^2A=1)
.'.(1/cosA)*(1/sinA)=1/(cosA.sinA)=secA.cosecA..
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