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Find the value of sec45 geometrically. Hence find sin 45 and cos 45 ?

$c o n s i d e r a r i g h t a n g l e t r i a n g l e t r i a n g l e, A B C w i t h r i g h t a n g l e a t B, s u c h t h a t, ∠ A = 45^{0} s o, ∠ A + ∠ B + ∠ C = 180^{0} \Rightarrow 45^{0} + 90^{0} + ∠ C = 180^{0} \Rightarrow ∠ C = 45^{0} t h e r e f o r e A B = B C {∠ A = ∠ C} S O, b y p y t h a g o r a s t h e o r e m, {A C}^{2} = {B C}^{2} + {C A}^{2} \Rightarrow A C = \sqrt{2} a s o ., \sec θ = \frac{h y p o t e n u s}{b a s e} = \frac{A C}{C B} = \frac{\sqrt{2} a}{a} = \sqrt{2} = \sec 45^{0} a s, \cos θ = \frac{1}{\sec θ} = \frac{1}{\sqrt{2}} = \cos 45^{0} s i m l a r l y, \sin θ = \frac{p e r p e n d i c u l a r}{h y p o t e n u s} = \frac{a}{\sqrt{2} a} = \frac{1}{\sqrt{2}} = \sin 45^{0}$

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