In the following figure, $t a n^{2} A - s e c^{2} A$

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Solution

The correct option is C $- 1$
Given, $△ A B C$ , A perpendicular from B on AC, let it cut AC at D, such that $∠ B D A = ∠ B D C = 90^{\circ}$
$A D = 3$
$B D = 4$
$B C = 12$
In $△ A B D$ ,
$A B^{2} = B D^{2} + A D^{2}$
$A B^{2} = 3^{2} + 4^{2}$
$A B = 5$
In $△ B C D$
$B C^{2} = C D^{2} + B D^{2}$
$12^{2} = C D^{2} + 4^{2}$
$C D^{2} = 128$
$C D = 8 \sqrt{2}$
Now, ${tan}^{2} A - {sec}^{2} A = (\frac{P}{B})^{2} - (\frac{H}{B})^{2} = (\frac{4}{3})^{2} - (\frac{5}{3})^{2}$
= $\frac{16}{9} - \frac{25}{9}$
= $\frac{16 - 25}{9}$
= $\frac{- 9}{9}$
= $- 1$

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