Question

How do you prove: `("cosec"(x - B))/ sec(x + B) = (1 - tan x tan B) / (tan x - tan B)` ?

Answer 1

We seek to prove the identity:

`("cosec"(x - B))/ sec(x + B) = (1 - tan x tan B) / (tan x - tan B)`

Consider the LHS:

`LHS = ("cosec"(x - B))/ sec(x + B)`

Using the fundamental definitions of trigonometric functions:

`LHS = (1/sin(x - B))/ (1/cos(x + B))`

`\ \ \ \ \ \ \ \ = (cos(x + B))/(sin(x - B))`

Now we use the sine and cosine sum of multiple angle identities:

`cos(A+B) -= cosAcosB - sinAsinB`
`sin(A+B) -= sinAcosB + cosAsinB`

So that:

`LHS = (cosxcosB-sinxsinB)/(sinxcosB-cosxsinB)`

`\ \ \ \ \ \ \ \ = (cosxcosB-sinxsinB)/(sinxcosB-cosxsinB) * (1/(cosxcosB))/(1/(cosxcosB))`

`\ \ \ \ \ \ \ \ = ((cosxcosB)/(cosxcosB) - (sinxsinB)/(cosxcosB) ) /( (sinxcosB)/(cosxcosB) - (cosxsinB)/(cosxcosB) )`

`\ \ \ \ \ \ \ \ = (1 - sinx/cosx * sinB/cosB ) /( sinx/cosx- sinB/cosB )`

`\ \ \ \ \ \ \ \ = (1 - tanx \ tanB ) /( tanx- tanB )`

`\ \ \ \ \ \ \ \ = RHS \ \ \ ` QED