# Establish identity3secangle/cscangle+sinangle/cosangle=4 tanangle

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You should come up with the substitution angle=`alpha` such that:

`(3 sec alpha)/(csc alpha) + sin alpha/cos alpha = 4 tan alpha`

You need to remember that `sec alpha` is reverse of `cos alpha ` such that:

`sec alpha = 1/cos alpha`

`` You need to remember that `csc alpha` is reverse of `sin alpha ` such that:

`csc alpha = 1/sin alpha`

Writing the left side in new form yields:

`(3/cos alpha)/(1/sin alpha) + sin alpha/cos alpha = 4 tan alpha`

You need to remember that tangent function is rational such that:

`tan alpha = sin alpha/cos alpha`

`` `(3/cos alpha)*(sin alpha) +tan alpha= 4 tan alpha`

`3 tan alpha + tan alpha = 4 tan alpha`

**The last line establishes the identity `(3 sec alpha)/(csc alpha) + sin alpha/cos alpha = 4 tan alpha.` **