Answer:

Step-by-step explanation:
I'm going to use the following identity to help with the difference inside the tangent function there:

Let
.
With some restriction on
this means:

We need to find
.
is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.


Subtract 4/9 on both sides:

Take the square root of both sides:


The cosine value is positive because
is a number between
and
because that is the restriction on sine inverse.
So we have
.
This means that
.
Multiplying numerator and denominator by 3 gives us:

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

Let's continue on to letting
.
Let's go ahead and say what the restrictions on
are.
is a number in between 0 and
.
So anyways
implies
.
Let's use the Pythagorean Identity again I mentioned from before to find the sine value of
.



Subtract 1/49 on both sides:

Take the square root of both sides:




So since
is a number between
and
, then sine of this value is positive.
This implies:

So
.
Multiplying both top and bottom by 7 gives:
.
Let's put everything back into the first mentioned identity.


Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

