Question

prove this tanA/secA-1+tanA/secA+1=2cosecA plz make it if possible in photo form

Answer 1

Answer:

Step-by-step explanation:

Sorry can't be done by photo:

HS = tanA/secA - 1 + tanA/secA + 1

= sinA/cosA / 1/cosA - 1 + sinA/cosA / 1/cosA + 1 [because tan A = sin A / cos A and sec A = 1 / cos A]

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

= sin A / 1 - cos A + sin A / 1 + cos A ,

= [sin A (1 + cos A) + sin A (1 - cos A)] / (1 - cos A) (1 + cos A) ,

= [sin A + sin A cos A + sin A - sin A cos A] / 1 - cos(sq) A,

= 2 sin A / sin(sq) A [From identity: sin(sq) A + cos(sq) A = 1] ,

= 2 / sin A ,

= 2 x 1 / sin A ,

=2 co sec A.

Hope it helps you.

Answer 2

Answer:

LHS = tanA/secA - 1 + tanA/secA + 1

= sinA/cosA / 1/cosA - 1 + sinA/cosA / 1/cosA + 1 [because tan A = sin A / cos A and sec A = 1 / cos A]

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

= sinA / 1 - cosA + sinA / 1 + cos A

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

= [sinA + sinAcosA + sinA - sinAcosA] / 1 - cos(sq) A

= 2sinA / sin(sq) A [From identity: sin(sq) A + cos(sq) A = 1]

= 2 / sin A

= 2 x 1 / sinA

=2 cosecA

Step-by-step explanation: