Evaluate: (1 - tan 10°)/(1 + tan 10°)

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Another way of evaluating this expression is using the subtraction formula for tangent.  This is:

`tan(a-b)={tan a-tan b}/{1+tan a tan b}`

now let `a=45` and `b=10` , so `tan a=tan45=1` .  This means that the expression becomes:

`{1-tan10}/{1+tan10}`

`=tan(45-10)`

`=tan35`

`approx 0.700208`

This means that the expression is `tan35 approx...

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Another way of evaluating this expression is using the subtraction formula for tangent.  This is:

`tan(a-b)={tan a-tan b}/{1+tan a tan b}`

now let `a=45` and `b=10` , so `tan a=tan45=1` .  This means that the expression becomes:

`{1-tan10}/{1+tan10}`

`=tan(45-10)`

`=tan35`

`approx 0.700208`

This means that the expression is `tan35 approx 0.700208` .

Approved by eNotes Editorial Team
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You should substitute `tan 45^o`  for 1 such that:

`(1 - tan 10^o)/(1 + tan 10^o) = (tan 45^o - tan 10^o)/(tan 45^o + tan 10^o)`

You need to convert the difference and the sum into a product, such that:

`tan 45^o - tan 10^o = (sin(45^o - 10^o))/(cos 45^o cos 10^o)`

`tan 45^o+ tan 10^o = (sin(45^o + 10^o))/(cos 45^o cos 10^o)`

`(1 - tan 10^o)/(1 + tan 10^o) = ((sin(45^o - 10^o))/(cos 45^o cos 10^o))/((sin(45^o + 10^o))/(cos 45^o cos 10^o))`

Reducing duplicate factors yields:

`(1 - tan 10^o)/(1 + tan 10^o) = (sin(45^o - 10^o))/(sin(45^o + 10^o))`

`(1 - tan 10^o)/(1 + tan 10^o) = (sin35^o)/(sin55^o)=0.573/0.819 ~~ 0.7`

Hence, evaluating the given expression yields `(1 - tan 10^o)/(1 + tan 10^o) = (sin35^o)/(sin55^o) ~~ 0.7.`

Approved by eNotes Editorial Team