Question

The person will be considered as genius who will solve it

Answer 1

Multiplying (1+cosA) with both numerator and denominator,
sinA + sinAcosA/ (1+cosA) (1-cosA)
sinA + sinAcosA/ 1-cos^2 A
sinA + sinAcosA/ sin^2 A
sinA(1+cosA)/sinAsinA
= 1/sinA + cosA/sinA
=cosecA + cotA

Answer 2

       sin a 
---------------------  = cosec a + cot a
  1 - cos a 

By multiplying the numerator and denominator of L.H.S by ( 1 + cos a ).

   ( sin a ) ( 1 + cos a )
---------------------------------- = cosec a + cot a
  ( 1 - cos a ) ( 1 + cos a )

 
  sin a + sin a x cos a
------------------------------- = cosec a + cot a
       1 - cos^2 a

sin a + sin a x cos a
---------------------------- = cosec a + cot a  { ( 1 - cos^2 a = sin^2 a }
           sin^2 a

 sin a  +   sin a x cos a
--------      -----------------  = cosec a + cot a
 sin^2 a       sin^2 a

    1        +        cos a
 ------               --------- = cosec a + cot a
  sin a               sin a

 cosec a + cot a = cosec a + cot a 
                                                                                                                                             [ { 1/sin a = cosec a } , { ( cos a / sin a ) = cot a } ]

Proved.