Question

3. find the value of sec 45 geometrically​

Answer 1

Answer:

Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°

∴∠ACB will also be 45°

∵∠BAC = ∠ACB = 45°

∴ AB = BC

Let AB = x

Then BC = x (∵AB = BC)

By Pythagorean Theorem,

AC = √(AB² + BC²)

     = √(x² + x²)  (∵ AB = BC = x)

     =√(2x²)

∴ AC = √2 x

Now we know that secФ = hypotenuse/adjacent(base)

∴ cos∠BAC = AB/AC

or, sec45° = √2 x/x  (∵ ∠BAC = 45°)

∴ sec45° = √2

Answer 2

sec theta is the reciprocal of cos theta

theta=45°

$\sec(45) = \frac{1}{ \cos(45) } \\ \sec(45) = \frac{1}{ \frac{1}{ \sqrt{2} } } \\ \sec(45) = 1 \times \frac{ \sqrt{2} }{1} \\ \sec(45) = \sqrt{2}$

i hope this will surely help you mate....