If sin A = cos A, find the value of

2 tan² A - 2 sec² A + 5.

sin A = cos A

tan A = 1

tan A = $\dfrac{Perpendicular}{Base}$

i.e., $\dfrac{Perpendicular}{Base} = 1$

∴ Length of BC = AB = x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = x² + x²

⇒ AC² = 2x²

⇒ AC = $\sqrt{2\text{x}^2}$

⇒ AC = x$\sqrt{2}$

sec A = $\dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB} = \dfrac{x \sqrt{2}}{x} = \sqrt{2}$

Now, 2 tan² A - 2 sec² A + 5

$= 2 \times 1^2 - 2 \times (\sqrt{2})^2 + 5 \\[1em] = 2 \times 1 - 2 \times 2 + 5 \\[1em] = 2 - 4 + 5 \\[1em] = 3$

Hence, 2 tan² A - 2 sec² A + 5 = 3.