Question

1+cosA/1-cosA - 1-cos A/1+cosA=4cotAcosecA

Answer 1

Trigonometry:

Trigonometry identities used:

`\boxed{\bf Sin^2 \ \theta + Cos^2 \ \theta = 1}`

`\sf \text{Multiply the numerator and denominator of}\ (1 + Cos \ A )/(1 - Cos \ A) \ \text{by the conjugate of 1 - Cos A}`

Conjugate of 1 - Cos A = 1 + Cos A

`\text{\bf Multiply the numerator and denominator of} \ (1 - Cos \ A)/(1 + Cos A) \ \text{ by the conjugate of 1 + Cos A}`

Conjugate of 1 + Cos A = 1 - Cos A

`\sf (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))\\\\\\`

`\sf = ((1 + Cos \ A)^2)/(1-Cos^2 \ A) - ((1 - Cos \ A)^2)/(1-Cos^2 \ A)\\\\\\= ((1 + Cos \ A)^2-(1 - Cos \ A)^2)/(1-Cos^2 \ A)\\\\\\We \ know \ that \ (a + b)^2 - (a - b)^2 = 4ab. Here, \ a = 1 \ & \ b = Cos \ A\\\\\\= (4*1*Cos \ A)/(Sin^2 \ A)\\\\\\=(4Cos\ A)/(Sin^2 \ A)`

`\sf = 4 * (1)/(Sin \ A)*(Cos \ A)/(Sin \ A)\\\\\\=4Cosec \ A *Cot \ A = RHS`

Hence, proved.