Cos A minus Sin A + 1 / Cos A + Sin A minus 1 is equal to cosec A + cot a
Answers
Answered by
2
heya aa
it's easy so easy actually.
cosA-sinA+1/cos+sinA-1(dividing by sin¢ on numerator and denominator)
then,it will be ,
=)cot¢-1+cosec¢/cot¢+1-cosec¢
cot¢+cosec¢-(cosec^2¢-cot^2¢)/cot¢+1-cosec¢
=)(cosec¢+cot¢)*cot¢-cosec¢+1/cot¢+1-cosec¢
cot¢+1-cosec¢ cancelled .
then,cosec¢+cot¢ Rhs proove ..
.
hope it help you.
@rajukumar☺1
it's easy so easy actually.
cosA-sinA+1/cos+sinA-1(dividing by sin¢ on numerator and denominator)
then,it will be ,
=)cot¢-1+cosec¢/cot¢+1-cosec¢
cot¢+cosec¢-(cosec^2¢-cot^2¢)/cot¢+1-cosec¢
=)(cosec¢+cot¢)*cot¢-cosec¢+1/cot¢+1-cosec¢
cot¢+1-cosec¢ cancelled .
then,cosec¢+cot¢ Rhs proove ..
.
hope it help you.
@rajukumar☺1
Answered by
1
ANSWER
..........
LHS=cosA−sinA+1cosA+sinA−1
=sinA(cosA−sinA+1)sinA(cosA+sinA−1)
=sinAcosA−sin2A+sinAsinA(cosA+sinA−1)
=sinAcosA+sinA−(1−cos2A)sinA(cosA+sinA−1)
=sinA(cosA+1)−(1−cosA)(1+cosA)sinA(cosA+sinA−1)
=(1+cosA)(sinA+cosA−1)sinA(cosA+sinA−1)
=(1+cosA)(sinA+cosA−1)sinA(cosA+sinA−1)
=1sinA+cosAsinA
=cscA+cotA=RHS
..........
LHS=cosA−sinA+1cosA+sinA−1
=sinA(cosA−sinA+1)sinA(cosA+sinA−1)
=sinAcosA−sin2A+sinAsinA(cosA+sinA−1)
=sinAcosA+sinA−(1−cos2A)sinA(cosA+sinA−1)
=sinA(cosA+1)−(1−cosA)(1+cosA)sinA(cosA+sinA−1)
=(1+cosA)(sinA+cosA−1)sinA(cosA+sinA−1)
=(1+cosA)(sinA+cosA−1)sinA(cosA+sinA−1)
=1sinA+cosAsinA
=cscA+cotA=RHS
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