Question

(1+cotA+tanA)(sinA-CosA)=secA/cosec²A - cosecA/sec²A = sinAtanA -cotA cosA

Answer 1

LHS= (1+cotA+tanA)(sinA-cosA)
=sinA+sinAcotA+sinAtanA-cosA-cosAcotA-
cosAtanA
=sinA+sinAcosA/sinA+sinAsinA/cosA-cosA
-cosAcosA/sinA-cosAsinA/cosA
=sinA+cosA+sin^2A/cosA-cosA-
cos^2A/sinA-sinA
=sin^2A/cosA-cos^2A/sinA
=sinA(sinA/cosA)-cosA(cosA/sinA) ----(1)
=1/cosecA(1/cosecA/1/secA)-1/secA(1/secA/
1/cosecA)
=(1/cosec^2A)(secA)-(1/sec^2)(cosecA)
=secA/cosec^2A-cosecA/sec^2A ----(2)

=sinAtanA-cosAcotA ---------------(3)
from (1)
from 2 and 3
=RHS

Answer 2

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$\it\huge\mathfrak\red{Hello}$]

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$<b>$]

LHS= (1+cotA+tanA)(sinA-cosA)

=sinA+sinAcotA+sinAtanA-cosA-cosAcotA-

cosAtanA

=sinA+sinAcosA/sinA+sinAsinA/cosA-cosA

-cosAcosA/sinA-cosAsinA/cosA

=sinA+cosA+sin^2A/cosA-cosA-

cos^2A/sinA-sinA

=sin^2A/cosA-cos^2A/sinA

=sinA(sinA/cosA)-cosA(cosA/sinA) ----(1)

=1/cosecA(1/cosecA/1/secA)-1/secA(1/secA/

1/cosecA)

=(1/cosec^2A)(secA)-(1/sec^2)(cosecA)

=secA/cosec^2A-cosecA/sec^2A ----(2)

=sinAtanA-cosAcotA ---------------(3)

from 2 and 3

LHS =RHS

$\huge\boxed\text{\fcolourbox{red}{aqau}{LHS=RHS}}}$]

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$\it\huge\mathfrak\red{thanks}$]

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