Question

find the value of theta, if
$\tan \: theta \: + \cot \: theta \: = 4 \div \sqrt{3}$
​

Answer 1

Answer:-

Given:-

tan θ + cot θ = 4/√3

Using tan θ = sin θ/cos θ and cot θ = cos θ/sin θ we get;

⟹ (sin θ/cos θ) + (cos θ/sin θ) = 4/√3

Now, taking LCM in LHS we get,

⟹ (sin² θ + cos² θ) / cos θ sin θ = 4/√3

using the identity sin² θ + cos² θ = 1 we get,

⟹ 1/cos θ sin θ = 4/√3

Now,

On squaring both sides we get,

⟹ 1/sin² θ cos² θ = (4/√3)²

Using cos² θ = 1 - sin² θ we get,

⟹ 1/ sin² θ (1 - sin² θ) = 16/3

⟹ 1/sin² θ - sin⁴ θ = 16/3

On cross multiplication we get,

⟹ 3 = 16(sin² θ - sin⁴ θ)

⟹ 3 = 16 sin² θ - 16 sin⁴ θ

⟹ 16 sin⁴ θ - 16 sin² θ + 3 = 0

⟹ 16 sin⁴ θ - 12 sin² θ - 4 sin² θ + 3 = 0

⟹ 4 sin² θ (4 sin² θ - 3) - 1(4 sin² θ - 3) = 0

⟹ (4 sin² θ - 1)(4 sin² θ - 3) = 0

★ 4 sin² θ - 1 = 0

⟹ 4 sin² θ = 1

⟹ sin² θ = (1/4)

⟹ (sin θ)² = (1/2)²

⟹ sin θ = 1/2

• sin 30° = 1/2

So,

⟹ θ = 30°

(or)

★ 4 sin² θ - 3 = 0

⟹ 4 sin² θ = 3

⟹ sin² θ = (3/4)

⟹ (sin θ)² = (√3/2)²

⟹ sin θ = √3/2

• sin 60° = √3/2.

⟹ θ = 60°

∴ The possible values of " θ " are 30° and 60°.

Answer 2

Given : $\sf tan \: theta \: + cot \: theta \: = \dfrac{4 }{ \sqrt{3}}$

Exigency To Find : The Value of $\theta$ [ theta ]

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⠀⠀⠀⠀⠀Finding value of $\theta$ [ theta ] :

$\qquad :\implies \sf tan \: theta \: + cot \: theta \: = \dfrac{4 }{ \sqrt{3}}$

$\dag\:\:\it{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ tan \: \theta \: = \: \dfrac{sin \:\theta }{cos \theta } }\bigg\rgroup$

$\qquad :\implies \sf \:\dfrac{sin \:\theta }{cos \theta } + cot \: theta \: = \dfrac{4 }{ \sqrt{3}}$

$\dag\:\:\it{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ cot \: \theta \: = \: \dfrac{cos \:\theta }{sin \theta } }\bigg\rgroup$

$\qquad :\implies \sf \:\dfrac{sin \:\theta }{cos \theta } + \dfrac{cos \:\theta }{sin \theta } \: = \dfrac{4 }{ \sqrt{3}}$

$\qquad :\implies \sf \:\dfrac{sin^2 \:\theta + cos^2 \theta }{cos \theta sin \theta } \: = \dfrac{4 }{ \sqrt{3}}$

$\dag\:\:\it{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ sin^2 \theta + cos^2 \theta = 1 }\bigg\rgroup$

$\qquad :\implies \sf \:\dfrac{1 }{cos \theta sin \theta } \: = \dfrac{4 }{ \sqrt{3}}$

By Squaring Both L.H.S & R.H.S :

$\qquad :\implies \sf \:\dfrac{1^2 }{cos^2 \theta sin^2 \theta } \: =\bigg( \dfrac{4 }{ \sqrt{3}}\bigg)$

$\qquad :\implies \sf \:\dfrac{1 }{cos^2 \theta sin^2 \theta } \: = \dfrac{16 }{ 3}$

$\dag\:\:\it{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ 1- sin^2 \theta cos^2 \theta }\bigg\rgroup$

$\qquad :\implies \sf \:\dfrac{1 }{(1 - sin^2 \theta ) sin^2 \theta } \: = \dfrac{16 }{ 3}$

$\qquad :\implies \sf \:\dfrac{1 }{ sin^2 \theta - sin^4 \theta } \: = \dfrac{16 }{ 3}$

$\qquad :\implies \sf \:16( sin^2 \theta - sin^4 \theta ) \: = 3$

$\qquad :\implies \sf \:16 sin^2 \theta - 16sin^4 \theta \: = 3$

$\qquad :\implies \sf \:16 sin^2 \theta - 16sin^4 \theta +3 \: = 0$

$\qquad :\implies \sf \: 16sin^4 \theta - 12sin^2 \theta - 4sin^2\theta +3 \: = 0$

$\qquad :\implies \sf \: 4sin^2 \theta ( 4sin^2 \theta - 3) -1 ( 4sin^2\theta - 3) \: = 0$

$\qquad :\implies \sf \: ( 4sin^2 \theta - 1 ) ( 4sin^2\theta - 3) \: = 0$

Therefore,

$\qquad :\implies \sf \: 4sin^2 \theta - 1 \: = 0$

$\qquad :\implies \sf \: 4sin^2 \theta \: = 1$

$\qquad :\implies \sf \: sin^2 \theta \: = \dfrac{1}{4}$

$\qquad :\implies \sf \: sin^2 \theta \: = \bigg(\dfrac{1}{2}\bigg)^2$

$\qquad :\implies \sf \: sin \theta \: = \dfrac{1}{2}$

$\qquad \dag\:\:\bigg\lgroup \sf{ sin 30^\circ = \dfrac{1}{2} }\bigg\rgroup$

$\qquad :\implies \bf \: \theta \: = 30^\circ$

Either ,

$\qquad :\implies \sf \: 4sin^2 \theta - 3 \: = 0$

$\qquad :\implies \sf \: 4sin^2 \theta \: = 3$

$\qquad :\implies \sf \: sin^2 \theta \: = \dfrac{3}{4}$

$\qquad :\implies \sf \: sin^2 \theta \: = \bigg(\dfrac{\sqrt{3}}{2}\bigg)^2$

$\qquad :\implies \sf \: sin \theta \: = \dfrac{\sqrt{3}}{2}$

$\qquad \dag\:\:\bigg\lgroup \sf{ sin 30^\circ = \dfrac{\sqrt{3}}{2} }\bigg\rgroup$

$\qquad :\implies \bf \: \theta \: = 60^\circ$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Values \:of\:\theta \:[theta] \:can\:be\:\bf{30^\circ \:\& \:60^\circ}}}}$

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