Question

Question no 15 PLZ solve ASAP

Answer 1

HOLAA MATEY :))

LHS

= sinA/1+cosA + 1+cosA/sinA

= [sin²A + (1 + cosA)²]/sinA(1 + cosA)

= (sin²A + 1 + cos²A + 2cosA)/sinA(1 + cosA)

= (1 + 1 + 2cosA)/sinA(1 + cosA) [since,sin²A + cos²A = 1]

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

= 2/sinA = 2 cosecA [since 1/sinA = cosecA]

= RHS

HENCE PROVED !!!

WISH IT HELPS `-`

~BEBRAINLY~

Answer 2

Step by explanation:

LHS :

cosA/1+sinA + 1+sinA/cosA

cos²A+1+sin²A+2sinA/1+sinA·cosA

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

2+2sinA/(1+sinA)cosA

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

[ (1+sinA) both will get reduce]

2/cosA= 2secA  

⇒ RHS

PLZ MARK IT AS BRAINLIST!!