IN THE WORLD OF THETA

max value of cos must be 1 and t^2 is always +ve . so if put any +ve fraction value less than 1 the result will be greater than 1 which is not possible.

For $\cos { \theta } =\frac { 1+{ t }^{ 2 } }{ 1-{ t }^{ 2 } }$ to be valid,

$-1\le \frac { 1+{ t }^{ 2 } }{ 1-{ t }^{ 2 } } \le 1$

Solving $t$ in inequalities,

$\frac { 1+{ t }^{ 2 } }{ 1-{ t }^{ 2 } } \ge -1\\ \Rightarrow 1+{ t }^{ 2 }\ge -1+{ t }^{ 2 }\\ \Rightarrow 2\ge 0$

$\frac { 1+{ t }^{ 2 } }{ 1-{ t }^{ 2 } } \le 1\\ \Rightarrow 1+{ t }^{ 2 }\le 1-{ t }^{ 2 }\\ \Rightarrow 2{ t }^{ 2 }\le 0\\ \Rightarrow t=0$

We've shown that for $\cos { \theta } =\frac { 1+{ t }^{ 2 } }{ 1-{ t }^{ 2 } }$ to be valid, $t=0$ .

the value of t is zero