Question

Prove the following trigonometric identity.

`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`

Answer 1

Solving the function using trignometric identities:

As we have `(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`

LHS = `(sin theta - cos theta + 1)/(sin theta + cos theta - 1)`

Dividing the numerator and denomenator by cos θ

`(sin theta/cos theta - cos theta/cos theta + 1/cos theta)/(sin theta/cos theta + cos theta/cos theta - 1/cos theta)`

= `(tan theta - 1 + sec theta)/(tan theta + 1 - sec theta)`

Multiplying and dividing by (tan θ - sec θ),

= `(tan theta + sec theta - 1)/(tan theta - sec theta + 1)xx (tan theta - sec theta)/(tan theta - sec theta)`

[(tan θ + sec θ)(tan θ - sec θ = tan2θ - sec2θ)]

= `[((tan^2 theta - sec^2 theta) - (tan theta - sec theta))/((tan theta - sec theta + 1)(tan theta - sec theta))]`

Using the identity sec2θ - tan2 θ = 1,

= `((-1 - tan theta + sec theta))/([(tan theta - sec theta + 1)(tan theta - sec theta)])`

= `(-1)/(tan theta - sec theta)`

= `1/(sec theta - tan theta)`

= RHS Hence proved