Question

prove that sec∅/tan∅+cot∅/cosec∅=sin∅+cos∅​

Answer 1

Answer:

Let ∅ be 'A'

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)=> [(1 - sin²A)/sinA][(1 - cos²A)/cosA]

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)=> [(1 - sin²A)/sinA][(1 - cos²A)/cosA]=> (cos²A/sinA)(sin²A/cosA)

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)=> [(1 - sin²A)/sinA][(1 - cos²A)/cosA]=> (cos²A/sinA)(sin²A/cosA)=> sinAcosA/1

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)=> [(1 - sin²A)/sinA][(1 - cos²A)/cosA]=> (cos²A/sinA)(sin²A/cosA)=> sinAcosA/1=>(sinAcosA)/(sin²A + cos²A)

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)=> [(1 - sin²A)/sinA][(1 - cos²A)/cosA]=> (cos²A/sinA)(sin²A/cosA)=> sinAcosA/1=>(sinAcosA)/(sin²A + cos²A)=> 1/[(sin²A/sinAcosA) + (cos²A/sinAcosA)]

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)=> [(1 - sin²A)/sinA][(1 - cos²A)/cosA]=> (cos²A/sinA)(sin²A/cosA)=> sinAcosA/1=>(sinAcosA)/(sin²A + cos²A)=> 1/[(sin²A/sinAcosA) + (cos²A/sinAcosA)]=> 1/(tanA + cotA)

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)=> [(1 - sin²A)/sinA][(1 - cos²A)/cosA]=> (cos²A/sinA)(sin²A/cosA)=> sinAcosA/1=>(sinAcosA)/(sin²A + cos²A)=> 1/[(sin²A/sinAcosA) + (cos²A/sinAcosA)]=> 1/(tanA + cotA)=> RHS

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)=> [(1 - sin²A)/sinA][(1 - cos²A)/cosA]=> (cos²A/sinA)(sin²A/cosA)=> sinAcosA/1=>(sinAcosA)/(sin²A + cos²A)=> 1/[(sin²A/sinAcosA) + (cos²A/sinAcosA)]=> 1/(tanA + cotA)=> RHSLHS = RHS

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)=> [(1 - sin²A)/sinA][(1 - cos²A)/cosA]=> (cos²A/sinA)(sin²A/cosA)=> sinAcosA/1=>(sinAcosA)/(sin²A + cos²A)=> 1/[(sin²A/sinAcosA) + (cos²A/sinAcosA)]=> 1/(tanA + cotA)=> RHSLHS = RHSHence Proved

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)=> [(1 - sin²A)/sinA][(1 - cos²A)/cosA]=> (cos²A/sinA)(sin²A/cosA)=> sinAcosA/1=>(sinAcosA)/(sin²A + cos²A)=> 1/[(sin²A/sinAcosA) + (cos²A/sinAcosA)]=> 1/(tanA + cotA)=> RHSLHS = RHSHence ProvedHope this helps....:)

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)=> [(1 - sin²A)/sinA][(1 - cos²A)/cosA]=> (cos²A/sinA)(sin²A/cosA)=> sinAcosA/1=>(sinAcosA)/(sin²A + cos²A)=> 1/[(sin²A/sinAcosA) + (cos²A/sinAcosA)]=> 1/(tanA + cotA)=> RHSLHS = RHSHence ProvedHope this helps....:)Smenevacuundacy and 40 more users found this answer helpful

Let ∅ be 'A'Given => (cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)LHS=> (cosecA - sinA)(secA - cosA)=> (1/sinA - sinA)(1/cosA - cosA)=> [(1 - sin²A)/sinA][(1 - cos²A)/cosA]=> (cos²A/sinA)(sin²A/cosA)=> sinAcosA/1=>(sinAcosA)/(sin²A + cos²A)=> 1/[(sin²A/sinAcosA) + (cos²A/sinAcosA)]=> 1/(tanA + cotA)=> RHSLHS = RHSHence ProvedHope this helps....:)Smenevacuundacy and 40 more users found this answer helpfulTHANKS 

rosariomividaa3 and 31 more users found this answer helpful

Answer 2

Step-by-step explanation:

Reason: Lipases can break large fat droplets into smaller ones." ... Lipases can digest fat in significant amounts only when large fat droplets are broken into tiny droplets to form a fina emulsion. Emulsification of fats by bile salts thus increases the lipase action on fats.

Assertion (A): Lipase helps in emulsification of fats.Reason (R)

People also search for

bile helps in

emulsification of fats in

lipase hydrolyses

fats and oils

what helps in

emulsification of fats

16 assertion a lipases help in emulsification of

fats reason r lipases hydrolyses fats and oils

emulsification of fats

is done by which organ

assertion lipase help in emulsification of

fats reason lipases hydrolyses fats and oils

assertion lipase help in emulsification of

fats reason lipase hydrolyzes fats and oils

• emulsification of large fat droplets

into small globules is performed by