Find the value of sec θ.

(a) From the figure (1) given below, find the value of sec θ.

(b) From the figure (2) given below, find the values of:

Answer :

(a) From the figure, Sec θ = AB/BD

But in ∆ADC, ∠D = 90^o

AC² =AD² + DC² 

⇒ (13)² =AD² + 25

⇒ AD² = 169 -25

= 144

= (12)²

⇒ AD = 12

AB² = AD² + BD² (in right ∆ABD)

= (12)² + (16)²

= 144 + 256

= 400

= (20)²

⇒ AB = 20

Sec θ = AB/BD

= 20/16

= 5/4

(b) let given ∆ABC

BD = 3, AC = 12, AD = 4

In right angled ∆ABD

AB² =AD² + BD²

⇒ AB² = (4)² + (3)²

⇒ AB² = 16 + 9

⇒ AB² = 25

⇒ AB² = (5)²

⇒ AB = 5

In right angled triangle ACD

AC² = AD² + CD²

⇒ CD² = AC² – AD²

⇒ CD² = (12)² – (4)²

⇒ CD² = 144 – 16

⇒ CD² = 128

⇒ CD = √128

⇒ CD = √64 × 2 CD

= 8√2

(i) sin x = perpendicular/Hypotenuse

= AD/AB

= 4/5

(ii) cot x = Base/Perpendicular

= BD/AD

= ¾

(iii) cot x = Base/Perpendicular

BD/AD

= 3/4

(iv) cosec x = Hypotenuse/Perpendicular

AB/BD

= 5/4

cot² x – cosec² x

= (3/4)² – (5/4)²

= 9/16 – 25/16

(9 -25)/16

= -16/16

= -1

Perpendicular = Hypotenuse/Base (in right angled ∆ACD)

= AD/CD

= 12/(8 √2)

= 3/(2 √2)

cot y = Base/Hypotenuse

= AD/CD

= 4/8 √ 2

= 1/2 √2

cot y = Base/Hypotenuse (in right angled ∆ACD)

= CD/AC

= 8√2/12

= 2√/3

Now tan² y = 1/cos² y

= (1/2√2)² – 1/(2√2/3)²

= ¼ × – ¼ × 2

= (1/8) – (9/8)

= (1-9)/8

= -8/8

= -1

tan² y – 1/cos² y = –1

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