Question

How do you prove `\frac { \cos a } { 1- \sin a } = \sec a + \tan a`?

Answer 1

Use `secx = 1/cosx` and `tanx = sinx/cosx`.

`(cosa)/(1 - sina) = 1/cosa + sina/cosa`

`cosa/(1 - sina) = (1 + sina)/cosa`

Multiply both sides by `cosa`.

`cosa/(1 - sina)cosa = (1 + sina)/cosa (cosa)`

`cos^2a/(1 - sina) = 1 + sina`

Now apply `sin^2x+ cos^2x = 1`.

`(1 - sin^2a)/(1 - sina) = 1 + sina`

`((1 + sina)(1 - sina))/(1 - sina) = 1 + sina`

`1 + sina = 1 + sina`

Since both sides are obviously equal, we've proved this identity.

Hopefully this helps!

Answer 2

see explanation.

Explanation:

Rearrange the left side or the right side until it is equivalent to the other side.

`"left side "=cosalpha/(1-sinalpha)`

`"multiply numerator/denominator by " 1+sinalpha"`

`color(white)(left side)=cosalpha/(1-sinalpha)xx(1+sinalpha)/(1+sinalpha)`

`color(white)(left side)=(cosalpha+cosalphasinalpha)/(1-sin^2alpha)`

`[• cos^2alpha+sin^2alpha=1to1-sin^2alpha=cos^2alpha]`

`color(white)(left side)=cosalpha/(cos^2alpha)+(cosalphasinalpha)/(cos^2alpha)`

`color(white)(left side)=1/cosalpha+sinalpha/cosalpha`

`color(white)(left side)=secalpha+tanalpha=" right side"to" verified"`