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

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

(i) sin x

(ii) cot x

(iii) cot^{2} x- cosec^{2} x

(iv) sec y

(v) tan^{2} y – 1/cos^{2} y.

**Answer :**

**(a)** From the figure, Sec θ = AB/BD

But in ∆ADC, ∠D = 90^{o}

AC^{2} =AD^{2 }+ DC^{2 }

⇒ (13)^{2} =AD^{2 }+ 25

⇒ AD^{2 }= 169 -25

= 144

= (12)^{2}

⇒ AD = 12

AB^{2 }= AD^{2 }+ BD^{2 }**(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

AB^{2} =AD^{2} + BD^{2}

⇒ AB^{2 }= (4)^{2} + (3)^{2}

⇒ AB^{2 }= 16 + 9

⇒ AB^{2} = 25

⇒ AB^{2} = (5)^{2}

⇒ AB = 5

In right angled triangle ACD

AC^{2} = AD^{2 }+ CD^{2}

⇒ CD^{2} = AC^{2} – AD^{2}

⇒ CD^{2} = (12)^{2 }– (4)^{2}

⇒ CD^{2 }= 144 – 16

⇒ CD^{2} = 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

cot^{2} x – cosec^{2} 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 tan^{2} y = 1/cos^{2 }y

= (1/2√2)^{2} – 1/(2√2/3)^{2}

= ¼ × – ¼ × 2

= (1/8) – (9/8)

= (1-9)/8

= -8/8

= -1

tan^{2 }y – 1/cos^{2 }y = –1

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