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Mathematics

If 5 cos θ = 6 sin θ; evaluate :

(i) tan θ

(ii) 12 sin θ−3 cos θ12 sin θ+3 cos θ\dfrac{\text{12 sin θ} - \text{3 cos θ}}{\text{12 sin θ} + \text{3 cos θ}}

Trigonometric Identities

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Answer

(i) Given:

5 cos θ = 6 sin θ

⇒ sin θcos θ=56\dfrac{\text{sin θ}}{\text{cos θ}} = \dfrac{5}{6}

⇒ tan θ=56\text{tan θ} = \dfrac{5}{6}

Hence, tan θ = 56\dfrac{5}{6}.

(ii) tan θ=56\text{tan θ} = \dfrac{5}{6}

tan θ = PerpendicularBase=56\dfrac{Perpendicular}{Base} = \dfrac{5}{6}

If 5 cos θ = 6 sin θ; evaluate : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

∴ If length of BC = 5x unit, length of AB = 6x unit.

In Δ ABC,

⇒ AC2 = BC2 + AB2 (∵ AC is hypotenuse)

⇒ AC2 = (5x)2 + (6x)2

⇒ AC2 = 25x2 + 36x2

⇒ AC2 = 61x2

⇒ AC = 61x2\sqrt{61\text{x}^2}

⇒ AC = 61\sqrt{61} x

sin θ = PerpendicularHypotenuse\dfrac{Perpendicular}{Hypotenuse}

=BCAC=5x61x=561= \dfrac{BC}{AC} = \dfrac{5x}{\sqrt{61} x} = \dfrac{5}{\sqrt{61}}

cos θ = BaseHypotenuse\dfrac{Base}{Hypotenuse}

=ABAC=6x61x=661= \dfrac{AB}{AC} = \dfrac{6x}{\sqrt{61} x} = \dfrac{6}{\sqrt{61}}

Now,

12 sin θ−3 cos θ12 sin θ+3 cos θ=12×561−3×66112×561+3×661=6061−18616061+1861=60−186160+1861=42617861=42617861=4278=713\dfrac{\text{12 sin θ} - \text{3 cos θ}}{\text{12 sin θ} + \text{3 cos θ}}\\[1em] = \dfrac{12 \times \dfrac{5}{\sqrt{61}} - 3 \times \dfrac{6}{\sqrt{61}}}{12 \times \dfrac{5}{\sqrt{61}} + 3 \times \dfrac{6}{\sqrt{61}}}\\[1em] = \dfrac{\dfrac{60}{\sqrt{61}} - \dfrac{18}{\sqrt{61}}}{\dfrac{60}{\sqrt{61}} + \dfrac{18}{\sqrt{61}}}\\[1em] = \dfrac{\dfrac{60 - 18}{\sqrt{61}}}{\dfrac{60 + 18}{\sqrt{61}}}\\[1em] = \dfrac{\dfrac{42}{\sqrt{61}}}{\dfrac{78}{\sqrt{61}}}\\[1em] = \dfrac{\dfrac{42}{\cancel{\sqrt{61}}}}{\dfrac{78}{\cancel{\sqrt{61}}}}\\[1em] = \dfrac{42}{78}\\[1em] = \dfrac{7}{13}

Hence, 12 sin θ−3 cos θ12 sin θ+3 cos θ=713\dfrac{\text{12 sin θ} - \text{3 cos θ}}{\text{12 sin θ} + \text{3 cos θ}} = \dfrac{7}{13}.

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