1 + 1/cosA=tan^2a/secA-1
plz prove it class 11th
Answers
Answered by
5
1+1/cosA (given)
1+secA. {1/cosA=secA}
{(secA+1)(secA-1)}/(secA-1)
(sec^2A-1)/(secA-1)
tan^2A/secA-1. hence proved
Hope it helps you ...✌✌
THANKS ☺☺
1+secA. {1/cosA=secA}
{(secA+1)(secA-1)}/(secA-1)
(sec^2A-1)/(secA-1)
tan^2A/secA-1. hence proved
Hope it helps you ...✌✌
THANKS ☺☺
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Answered by
6
Correction :
Prove that ,
1 + ( 1 / cos A ) = tan²A / ( sec A - 1 )
Here is your answer :
L.H.S = 1 + ( 1 / cos A )
R.H.S = tan²A / ( sec A - 1 )
______________________
Using identity,
=> tan A = sin A / cos A
=> sec A = ( 1 / cos A )
_______________________
= ( sin²A / cos²A ) / [ ( 1 / cos a ) - 1
= ( sin²A / cos²A ) / [ ( 1 - cos A ) / cos A ]
= ( sin²A / cos²A ) × [ cos A / ( 1 - cos A ) ]
= ( sin²A / cos A ) × [ 1 / ( 1 - cos A ) ]
= ( sin²A ) / [ cos A ( 1 - cos A ) ]
= ( sin²A ) / [ cos A - cos²A ]
_______________________
Using formula,
=> sin²A = ( 1 - cos²A )
________________________
= ( 1 - cos²A ) / [ cos A ( 1 - cos A ) ]
= [ ( 1 )² - ( cos A )² ] / [ cos A ( 1 - cos A ) ]
Using identity,
=> ( a² - b² ) = ( a + b ) ( a - b )
= [ ( 1 + cos A ) ( 1 - cos A ) ] / [ cos A ( 1 - cos A ) ]
= ( 1 + cos A ) / cos A
= ( 1 / cos A ) + ( cos A / cos A )
= ( 1 / cos A ) + 1
=> 1 + ( 1 / cos A ) = L.H.S
Proved !!
Prove that ,
1 + ( 1 / cos A ) = tan²A / ( sec A - 1 )
Here is your answer :
L.H.S = 1 + ( 1 / cos A )
R.H.S = tan²A / ( sec A - 1 )
______________________
Using identity,
=> tan A = sin A / cos A
=> sec A = ( 1 / cos A )
_______________________
= ( sin²A / cos²A ) / [ ( 1 / cos a ) - 1
= ( sin²A / cos²A ) / [ ( 1 - cos A ) / cos A ]
= ( sin²A / cos²A ) × [ cos A / ( 1 - cos A ) ]
= ( sin²A / cos A ) × [ 1 / ( 1 - cos A ) ]
= ( sin²A ) / [ cos A ( 1 - cos A ) ]
= ( sin²A ) / [ cos A - cos²A ]
_______________________
Using formula,
=> sin²A = ( 1 - cos²A )
________________________
= ( 1 - cos²A ) / [ cos A ( 1 - cos A ) ]
= [ ( 1 )² - ( cos A )² ] / [ cos A ( 1 - cos A ) ]
Using identity,
=> ( a² - b² ) = ( a + b ) ( a - b )
= [ ( 1 + cos A ) ( 1 - cos A ) ] / [ cos A ( 1 - cos A ) ]
= ( 1 + cos A ) / cos A
= ( 1 / cos A ) + ( cos A / cos A )
= ( 1 / cos A ) + 1
=> 1 + ( 1 / cos A ) = L.H.S
Proved !!
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