Prove that: 1+tanA/cosecA+1+cotA/secA=secA+cosecA
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Answered by
5
Hi ,
1 ) ( 1 + tanA )/cosecA
= ( 1 + sinA/cosA )/( 1 / sinA )
= [sinA ( cosA + sinA )]/cosA ---( 1 )
2 ) ( 1 + cot A )/secA
= ( 1 + cosA/sinA )/( 1/cosA )
=[ cosA ( sinA + cosA )]/sinA --( 2 )
Now ,
LHS = ( 1 ) + ( 2 )
=[sinA(sinA+cosA)]/cosA+[cosA(sinA+cosA)]/sinA
= ( sinA + cosA )[ sinA/cosA + cosA/sinA ]
= ( sinA + cosA )[ ( sin²A+cos²A)/sinAcosA ]
= ( sinA + cosA )/sinAcosA
= sinA/sinAcosA + cosA/sinAcosA
= 1/ cosA + 1/sinA
= secA + cosecA
= RHS
I hope this helps you.
: )
1 ) ( 1 + tanA )/cosecA
= ( 1 + sinA/cosA )/( 1 / sinA )
= [sinA ( cosA + sinA )]/cosA ---( 1 )
2 ) ( 1 + cot A )/secA
= ( 1 + cosA/sinA )/( 1/cosA )
=[ cosA ( sinA + cosA )]/sinA --( 2 )
Now ,
LHS = ( 1 ) + ( 2 )
=[sinA(sinA+cosA)]/cosA+[cosA(sinA+cosA)]/sinA
= ( sinA + cosA )[ sinA/cosA + cosA/sinA ]
= ( sinA + cosA )[ ( sin²A+cos²A)/sinAcosA ]
= ( sinA + cosA )/sinAcosA
= sinA/sinAcosA + cosA/sinAcosA
= 1/ cosA + 1/sinA
= secA + cosecA
= RHS
I hope this helps you.
: )
meenakshi310320:
Thank you
Answered by
3
Answer:
Step-by-step explanation:
(1).( 1 + tanA )/cosecA
= ( 1 + sinA/cosA )/( 1 / sinA )
= [sinA ( cosA + sinA )]/cosA ---( 1 )
(2).( 1 + cot A )/secA
= ( 1 + cosA/sinA )/( 1/cosA )
=[ cosA ( sinA + cosA )]/sinA --( 2 )
Now ,
LHS = ( 1 ) + ( 2 )
=[sinA(sinA+cosA)]/cosA+[cosA(sinA+cosA)]/sinA
= ( sinA + cosA )[ sinA/cosA + cosA/sinA ]
= ( sinA + cosA )[ ( sin²A+cos²A)/sinAcosA ]
= ( sinA + cosA )/sinAcosA
= sinA/sinAcosA + cosA/sinAcosA
= 1/ cosA + 1/sinA
= secA + cosecA
= RHS
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