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

1 Answer
Steve M
Apr 4, 2018

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

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