Plzz solve it for me. 1+1/cosA=tan^2A/secA-1.
Bagri22g1:
hoo
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Answered by
29
Hi ,
LHS = 1 + 1/cosA
[ Since 1/cosA = secA ]
= 1 + sec A
= [( Sec A + 1 )( sec A - 1 )]/(sec A - 1 )
= ( Sec² A - 1 ) /( secA - 1 )
= tan² A / ( secA - 1 )
[ Since sec² A - 1 = tan² A ]
= RHS
LHS = 1 + 1/cosA
[ Since 1/cosA = secA ]
= 1 + sec A
= [( Sec A + 1 )( sec A - 1 )]/(sec A - 1 )
= ( Sec² A - 1 ) /( secA - 1 )
= tan² A / ( secA - 1 )
[ Since sec² A - 1 = tan² A ]
= RHS
I have mistakenly rated your answer actually its correct
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4
this is the answer....
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It is wrong., Plz check
Yes it's incorrect
U have changed the question
But The question is correct
Any way u tried so thats why thx
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