Prove the following trigonometrical identity- 1. √(1+cos A)/(1-cos A) + √(1-cos A)/(1+cos A) = 2 cosec A 2. tan A/(1- cot A) + cot A/(1-...

1. √(1+cos A)/(1-cos A) + √(1-cos A)/(1+cos A) = 2 cosec A

L.H.S=√(1+cos A)/(1-cos A) + √(1-cos A)/(1+cos A)

=√(1+cos A)(1+cos A)/(1-cos A)(1+cos A) + √(1-cos A)(1-cos A)/(1+cos A)(1-cos A)

=(1+cos A)/(1-cos^2 (A)) + (1-cos A)/(1-cos^2 (A))

=(1+cos A)/sin^2(A) + (1-cos A)/sin^2(A)

=(1+cos A+1-cos A)/sin^2(A)

=2/sin^2(A)

=2csc^2(A)

if your question is √{(1+cos A)/(1-cos A)} + √{(1-cos A)/(1+cos A)}

=√{(1+cos A)/(1-cos A)} + √{(1-cos A)/(1+cos A)}

=√(1+cos A)(1+cos A)/√(1-cos A)(1+cos A) + √(1-cos A)(1-cos A)/√(1+cos A)(1-cos A)

=(1+cos A)/√(1-cos^2 (A)) + (1-cos A)/√(1-cos^2 (A))

=(1+cos A)/√sin^2(A) + (1-cos A)/√sin^2(A)

=(1+cos A+1-cos A)/sin(A)

= 2 cosec A

2. tan A/(1- cot A) + cot A/(1- tan A) = 1+ tan A + cot A

L.H.S=tan A/(1- cot A) + cot A/(1- tan A)

=[sinA/cosA]/[(sinA-cosA)/sinA] + [cosA/sinA]/[(cosA-sinA)/cosA]

=sin^2(A)/[(sinA-cosA)cosA] + cos^2(A)/[(cosA-sinA)sinA]

={sin^3(A)-cos^3(A)}/[(sinA-cosA)sinAcosA]

={(sinA-cosA)(sin^2(A)+sinAcosA+cos^2(A))}/[(sinA-cosA)sinAcosA]

=(sin^2(A)+sinAcosA+cos^2(A))/sinAcosA

=tanA+1+cotA

= 1+ tan A + cot A

=R.H.S