**Question 1. ****(a) From the figure (1) given below, find the values of:**

(i) sin θ

(ii) cos θ

(iii) tan θ

(iv) cot θ

(v) sec θ

(vi) cosec θ

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

(i) sin

(ii) cos A

(iii) sin^{2} A + cos^{2} A

(iv) sec^{2} A – tan^{2} A.

**Answer :**

**(a)** From right angled triangle OMP,

By Pythagoras theorem,

OP^{2} = OM^{2} +MP^{2}

⇒ MP^{2} = OP^{2} + OM^{2}

⇒ MP^{2} = (15)^{2} – (12)^{2}

⇒ MP^{2} = 225 – 144

⇒ MP^{2} = 81

⇒ MP^{2} = 9^{2}

⇒ MP = 9

**(i)** sin θ = MP/OP

= 9/15

= 3/5

**(ii)** cos θ = OM/OP

= 12/15

= 4/5

**(iii)** tan θ = MP/OP

= 9/12

= ¾

**(iv)** cot θ = OM/MP

= 12/9

= 4/3

**(v)** sec θ = OP/OM

= 15/12

= 5/4

**(vi)** cosec θ = OP/MP

= 15/9

= /3

**(b)** From right angled triangle ABC,

By Pythagoras theorem,

AB^{2} = AC^{2} + BC^{2}

⇒ AB^{2} = (12)^{2} + (5)^{2}

⇒ AB^{2} = 144 + 25

⇒ AB^{2} = 169

⇒ AB^{2} = 13^{2}

⇒ AB = 13

**(i)** sin A = BC/AB

= 5/13

**(ii)** cos A = AC/AB

= 12/13

**(iii)** sin^{2} A + cos^{2} A = (BC/AB)^{2} + (AC/AB)^{2}

= (5/13)^{2} + (12/13)^{2}

= (25/169) + (144/169)

= (25 + 144)/ 169

= 169/169

= 1

sin^{2} A + cos^{2} A = 1

**(iv)** sec^{2} A – tan^{2} A = (AB/AC)^{2} – (BC/AC)^{2}

= (13/12)^{2} – (5/12)^{2}

= (169/144) – (25/144)

= (169 – 25)/144

= 144/144

= 1

sec^{2} A – tan^{2} A = 1

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