Question

1+ cot2a/1+cosec a = cosec a​

Answer 1

To prove:

$l.h.s =1 + \frac{cot {}^{2} a}{1 + cosec \: a}$

$\: \: \: \: \: \: \: \: \: \: = \frac{1 + cosec \: a + cot {}^{2} }{1 + csec \: a}$

$\: \: \: \: \: \: \: \: \: \: = \frac{1 + cosec \: a + (cosec {}^{2} a - 1)}{1 + cosec \: a}$

[cot^2 a=(cosec^2 a-1)]

Now,

$\: \: \: \: \: \: \: \: \: \: = \frac{1 + cosec \: a + cosec {}^{2} \: a - 1}{1 + cosec \: a}$

$\: \: \: \: \: \: \: \: \: \: = \frac{cosec \: a(1 + cosec \: a)}{(1 + cosec \: a)}$

$\: \: \: \: \: \: \: \: \: \: = cosec \: a \: \: r.h.s \: proved$

Hence,Its is proved.

Answer 2

Answer:

I think this question is like cot²a

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]