# 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- tan A) = 1+ tan A + cot A for typing the square...

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- tan A) = 1+ tan A + cot A

for typing the square root sign plz press alt+251 on the num pad.

neela | High School Teacher | (Level 3) Valedictorian

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1)

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

Mutiply the first expression within the square root by 1+cosA both numerator and denominator.Similarly mutiply the second expression within the square root by 1+cosA both numerator and denominator.

√{(1+cos A)^2/[(1-cos A)(1+cosA)]} + √{(1-cos A)^2/[(1+cos A)(1-cosA)]}

(1-cosa)/sqrt(1-cos^2A) +(1+cosA)/sqrt(1-cos^2A)

(1-cosA)/sinA + 1+cosA)/sinA = 2/sinA = 2 cosecA = RHS

2)

tan A/(1- cot A) + cot A/(1- tan A)

= tanA/(1-1/tanA)  + 1/[TanA(1-tanA)]

=tan^2/(tanA-1)  - /[tanA(tanA-1)]

= (tan^3 A - 1)/{tanA(tanA-1).}

= (tanA -1)(tan^2+tanA+1)/ [tanA(tanA-1)], as a^3-b^3 =(a-b)(a^2+ab+b^2).

=(tan^2+tanA+1)/tanA, as tanA-1 in numerator and denominator gets cancelled.

=tanA+1+cotA = RHS

jeyaram | Student, Undergraduate | (Level 1) Valedictorian

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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

jeyaram | Student, Undergraduate | (Level 1) Valedictorian

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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