Question

Want the solution for the sum ....

Answer 1

putting value of k

(Cosec + cot )² - 1 ÷ (cosec + cot )² + 1

putting cosec = 1/sin and cot = cos/sin

(1/ sin + cos/sin ) ² - 1 ÷ (1/sin + cos/sin)² + 1

{(1 + cos)/ sin}² - 1 ÷ {(1 + cos)/ sin}² + 1
taking LCM
{(1 + cos² - sin²) ÷ sin²} ÷ {(1 + cos² + sin²) ÷ sin²

sin² will cut sin²

(1 + cos² - sin²) ÷ ( 1 + cos² + sin²)

putting 1 - sin² = cos² and

cos² + sin² = 1

(cos² + cos²) ÷ (1+ 1)

2cos² ÷ 2

cos² Hence proved

Answer 2

RHS=k²-1/k²+1
= (cosec thita+ cot thita)²-1/ (cosec thita+ cot thita)²+1
=cosec² thita+ cot²thita+2cosec thita cot thita -1/cosec²thita+cot² thita +2cosecthita cot thita
put 1= cosec² thita - cot² thita
= cosec²thita+cot²thita+2cosec thita cot thita -(cosec²thita-cot²thita)/cosec²thita+cot²thita+2cosec thita cot thita + (cosec²thita - cot²thita)
=after opening the bracket and canceling
=2cot²thita+2cosec thita cot thita/2cosec²thita+2 cosec thita cot thita
= 2cot thita(cot thita+cosec thita)/2cosec thita(cosec thita +cot thita)
=cot thita/cosec thita
=cos thita/sin thita*sin thita
=cos thita=LHS
Hope this will help u