Cosec(65+A) - sec(25-A)
Answers
Answered by
34
it can written as sec[90°-(65°+A)]-sec(25°-A)
= sec (90°-65°-A)-sec(25°-A)
=sec(25°-A)-sec(25°-A)
=0
= sec (90°-65°-A)-sec(25°-A)
=sec(25°-A)-sec(25°-A)
=0
Anonymous:
hope this helped u
Thank you so much
welcome
u can follow me that i can help in future also:)
Answered by
22
Hi ,
**********************************
We know that ,
Cosec ( 90 - A ) = sec A
***********************************
Cosec ( 65 + A ) = Cosec [(90-25 + A)]
= Cosec[ 90 -(25 - A )]
= Sec ( 25 - A )
According to the problem given ,
Cosec( 65 + A ) - sec ( 25 - A )
= Sec ( 25 - A ) - sec( 25 - A )
= 0
:)
**********************************
We know that ,
Cosec ( 90 - A ) = sec A
***********************************
Cosec ( 65 + A ) = Cosec [(90-25 + A)]
= Cosec[ 90 -(25 - A )]
= Sec ( 25 - A )
According to the problem given ,
Cosec( 65 + A ) - sec ( 25 - A )
= Sec ( 25 - A ) - sec( 25 - A )
= 0
:)
Similar questions