# Prove the identity: `(cos B + sin B)/(cos B - sinB) - (cos B - sinB)/(cosB + sin B) = 2 tan 2B`

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### 2 Answers

The identity `(cos B + sin B)/(cos B - sinB) - (cos B - sinB)/(cosB + sin B) = 2 tan 2B` has to be proved.

`(cos B + sin B)/(cos B - sinB) - (cos B - sinB)/(cosB + sin B)`

=> `((cos B + sin B)^2 - (cos B - sinB)^2)/(cos^2B - sin^2B)`

=> `(cos^2B + sin^2B + 2*sin B*cos B - cos^2B - sin^2B + 2*sin B*cos B)/(cos^2B - sin^2B)`

=> `(4*sin B*cos B)/(cos^2B - sin^2B)`

=> `(2*sin 2B)/cos 2B`

=> `2*tan 2B`

**This proves that ** `(cos B + sin B)/(cos B - sinB) - (cos B - sinB)/(cosB + sin B) = 2 tan 2B`

L:H:S ≡ (cosB + sin B)/(cosB - sinB) - (cosB - sinB)/(cosB + sinB)

= [(cosB + sinB)² - (cosB -sinB)²]/(cosB - sinB)(cosB + sinB)

**⇒ use in order : x² - Y² = (x+y)(x-y) ** **&** **(x+y)(x-y) = x² - Y² **

=(cosB+sinB+cosB-sinB)(cosB+sinB-cosB+sinB)/cos²B - sin²B

**⇒ use cos²A - sin²A = cos 2A**

= 2cosB.2sinB / cos2B

= 2 x 2sinB.cosB / cos2B

**⇒ use 2sinA.cosA = sin 2A**

= 2 x sin2B/ cos2B

= 2 tan2B

Therefore L:H:S ≡ R:H:S