cotA/1+tanA=cotA-1/2-sec squareA
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Answered by
2
Hi ,
LHS = cotA/( 1 + tanA )
= [CotA(1-tanA)]/[(1+tanA)(1-tanA)]
= [ cota-cotAtanA]/[ 1² - tan² A ]
= ( CotA - 1 ) / [ 1 - ( sec² A - 1 ) ]
= ( cotA - 1 )/( 1 - sec² A + 1 )
= ( cotA - 1 )/ ( 2 - sec² A )
= RHS
I hope this helps you.
: )
LHS = cotA/( 1 + tanA )
= [CotA(1-tanA)]/[(1+tanA)(1-tanA)]
= [ cota-cotAtanA]/[ 1² - tan² A ]
= ( CotA - 1 ) / [ 1 - ( sec² A - 1 ) ]
= ( cotA - 1 )/( 1 - sec² A + 1 )
= ( cotA - 1 )/ ( 2 - sec² A )
= RHS
I hope this helps you.
: )
Answered by
3
Hi,
Please see the attached file!
Thanks
Please see the attached file!
Thanks
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