If 1/cos - cos=a^3 and 1/sin - sin=b^3

Show that tan=a/b

a^3=1/cos-cos = (1-cos^2)/cos=sin^2/cos

b^3=1/sin-sin = (1-sin^2)/sin=cos^2/sin

a^3/b^3=sin^2/cos * cos^2/sin=sin^3/cos^3
=>
a/b=sin/cos = tan

(Hope you understand your own notations.)

To show that tan(a) = a/b, we need to start by manipulating the given equations and expressing them in terms of tan.

Starting with the first equation:
1/cos - cos = a^3

We can simplify this equation by taking cosine to the other side and getting:
1/cos = a^3 + cos

Now, we can rewrite 1/cos as sec:
sec = a^3 + cos

Next, we can rewrite sec in terms of sin:
1/cos = 1/sin*cos
1/cos = csc*sin*cos
1/cos = csc*sin

Since csc is the reciprocal of sin, we can rewrite it as:
1/cos = sin/sin*cos
1/cos = sin/cos

Finally, we have:
tan = sin/cos

Now, let's move on to the second equation:
1/sin - sin = b^3

We can simplify this equation by taking sine to the other side and getting:
1/sin = b^3 + sin

Now, we can rewrite 1/sin as csc:
csc = b^3 + sin

Next, we can rewrite csc in terms of cos:
1/sin = 1/cos*sin
1/sin = sec*cos*sin
1/sin = sec*sin

Since sec is the reciprocal of cos, we can rewrite it as:
1/sin = cos/cos*sin
1/sin = cos/sin

Finally, we have:
cot = cos/sin

Now, we have tat = sin/cos and cot = cos/sin. Considering the fundamental trigonometric identity tan(a) = 1/cot(a), we can see that:
tan = cot^-1
tan = (cos/sin)^-1
tan = sin/cos

Hence, we have shown that tan(a) equals a/b.