ANSWER
True
Step-by-step explanation
The given trigonometric equation is:
![{ \tan}^(2) x + 1 = { \sec}^(2) x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/58gg2henjrqr71buyxzeuypkhollbrcqwn.png)
We take the LHS and simplify to arrive at the RHS.
![{ \tan}^(2) x + 1 = \frac{{ \sin}^(2) x}{{ \cos}^(2) x} + 1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ozcujcu70r83gsez6rnomqufsnrwwozelb.png)
Collect LCM on the right hand side to get;
![{ \tan}^(2) x + 1 = \frac{{ \sin}^(2) x + {\cos}^(2) x}{{ \cos}^(2) x}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/bunzhim1dg8fo0huddvdqah944unzobfkz.png)
This implies that
![{ \tan}^(2) x + 1 = \frac{1}{{ \cos}^(2) x} .](https://img.qammunity.org/2020/formulas/mathematics/middle-school/umpetxpeu96l9hoz5av8nb6wnkeiscg8v0.png)
![{ \tan}^(2) x + 1 = {( (1)/( \cos(x) )) }^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/7p6me8n53mij1zn5ghijyuyrqnrcx11p7m.png)
![{ \tan}^(2) x + 1 = { \sec}^(2) x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/58gg2henjrqr71buyxzeuypkhollbrcqwn.png)
This identity has been verified .Therefore the correct answer is true.