**Given 4 sin θ = 3 cos θ, find the values of:**

(i) sin θ

(ii) cos θ

(iii) cot^{2} θ – cosec^{2} θ.

**Answer :**

4 sin θ = 3 cos θ

⇒ sin θ/cos θ = ¾

⇒ tan θ = ¾

Consider ∆ABC right angled at B and ∠ACB = θ

tan θ = perpendicular/base

¾ = AB/BC

⇒ AB/BC = ¾

Take AB = 3x then BC = 4x

In right angled ∆ABC

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

AC^{2} = (3x)^{2} + (4x)^{2}

AC^{2} = 9x^{2} + 16x^{2} = 25x^{2}

AC^{2} = (5x)^{2}

⇒ AC = 5x

**(i)** In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC = 3x/5x = 3/5

**(ii)** In right angled ∆ABC

cos θ = base/hypotenuse

cos θ = BC/AC = 4x/5x = 4/5

**(iii)** In right angled ∆ABC

cot θ = base/perpendicular

cot θ = BC/AB = 4x/3x = 4/3

In right angled ∆ABC

cosec θ = hypotenuse/perpendicular

cosec θ = AC/AB = 5x/3x = 5/3

cot^{2} θ – cosec^{2} θ = (4/3)^{2} – (5/3)^{2}

= 16/9 – 25/9

= (16 – 25)/9

= -9/9

= -1

Hence, cot^{2} θ – cosec^{2} θ = -1.

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