Question

plz help me with this question​

Answer 1

LHS = 1 + (Cot²A) / (1+CosecA)

= {1 +CosecA + Cot²A} / ( 1+ CosecA)

On multiplying Numerator & denominator by ( (1- CosecA)

= {( 1+CosecA +Cot²A) ( 1-CosecA)} / (1-Cosec²A)

= (1+CosecA+Cot²A-CosecA -Cosec²A -Cot²A.CosecA) / (1-Cosec²A)

Using fundamental trigonometric identity:

Cot²A+1 = Cosec²A , DENOMINATOR = 1-Cosec²A= -Cot²A

And NUMERATOR = 1+Cot²A-Cosec²A-Cot²A.CosecA

= Cosec²A-Cosec²A - Cot²A.CosecA

= -Cot²A.CosecA

Now, LHS = (-Cot²A.CosecA)/(-Cot²A)

= CosecA= RHS

[Hence Proved]

Deepansh Parmar

Hungenahalli Sitaramarao Badarinath

Answered June 8, 2017

1 +cot^2A/(1+cosec A) = cosec A

I presume the brackets as i have shown are correct as the question is incomplete without them.

LHS = 1 +cot^2A/(1+cosec A)

= 1 + ( cosec^2 -1)/[1+ cosec A]

= 1+ (cosec x+1)(cosec x -1)/[1+ cosec A]

= 1 + cosec x -1

= cosec x = RHS

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