Question

(1+tanA + secA) (1+cotA - cosecA)

Answer 1

Hi ,

i ) 1 + tanA + secA

= 1 + ( sinA/cosA ) + ( 1/cosA )

= ( cosA + sinA + 1 )/cosA ---( 1 )

ii ) 1 + cotA - Cosec A

= 1 + ( cosA/sinA ) - ( 1/sinA )

= ( sinA + cosA - 1 )/sinA ---( 2 )

According to the problem given ,

(1+tanA+secA)(1+cotA-cosecA)

= ( 1 ) × ( 2 )

=[(sinA+cosA+1)(sinA+cosA-1)]/(cosAsinA)

= [ ( sinA + cosA )² - 1² ]/( sinAcosA )

= ( sin² A + cos² A + 2sinAcosA -1)/(sinAcosA )

= ( 1 + 2sinAcosA - 1 )/( sinAcosA )

= ( 2sinAcosA )/ ( sinAcosA )

= 2

I hope this helps you.

: )

Answer 2

Hi,

Please see the attached file!


Thanks