Show That 1/(cosecA-cotA) - 1/sinA = 1/sinA - 1/(cosecA+cotA)

To prove  1/(cosecA-cotA) - 1/sinA = 1/sinA - 1/(cosecA+cotA)  we will start from LHS

Answer: LHS = RHS, 1/(cosecA-cotA) - 1/sinA = 1/sinA - 1/(cosecA+cotA) = cotA

Let us see detailed explanation

Explanation:

Consider  the LHS  1/(cosecA-cotA) - 1/sinA

=  1/ (1/sinA-cosA/sinA) - 1/sinA  [∵ cosec A = 1/sin A  and cot A = cos A /sin A ]

=1/ (1-cosA/sinA)-1/sinA  [Simplifying ]

=sinA / (1-cosA) - 1/sinA

=(sin²A - 1+cosA ) /sinA (1-cosA)

=-cos²A+cosA/sinA(1-cosA)

On taking (1-cosA) common from both numerator and denominator and simplifying it, we get

=cosA/sinA = cotA

Consider RHS = 1/sinA-1/(cosecA+cotA)

=sinA-1/(1/sinA+cosA/sinA)

= 1/sinA - 1/(1+cosA/sinA)

= 1/sinA-sinA/(1+cosA)

=1+cosA-sin²A/sinA(1+cosA)

= (1-sin²A)+cosA/sinA(1+cosA)

On taking (1+cosA) common from both numerator and denominator and simplifying it, we get

cosA/sinA = cot A

Hence proved

 LHS=RHS