Solve for x ∈ W, 0° x 90°. (i) 3 tan 2 2x = 1 (ii) tan 2 x = 3(sec x - 1)

(i) Given, ⇒ 3 tan 2 2x = 1 ⇒ tan 2 2x = ⇒ tan 2x = ⇒ tan 2x = ⇒ tan 2x = tan 30° ⇒ 2x = 30° ⇒ x = = 15°. Hence, x = 15°. (ii) Given, ⇒ tan 2 x = 3(sec x - 1) ⇒ sec 2 x - 1 = 3sec x - 3 ⇒ sec 2 x - 3 sec x - 1 + 3 = 0 ⇒ sec 2 x - 3 sec x + 2 = 0 ⇒ sec 2 x - 2 sec x - sec x + 2 = 0 ⇒ sec x(sec x - 2) - 1(sec x - 2) = 0 ⇒ (sec x - 1)(sec x - 2) = 0 ⇒ sec x - 1 = 0 or sec x - 2 = 0 ⇒ sec x = 1 or sec x = 2 ⇒ sec x = sec 0° or sec x = sec 60° ⇒ x = 0° or x = 60° Hence, x = 0° or 60°.

Mathematics

Solve for x ∈ W, 0° ≤ x ≤ 90°.
(i) 3 tan² 2x = 1
(ii) tan² x = 3(sec x - 1)

Trigonometric Identities

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Answer

(i) Given,

⇒ 3 tan² 2x = 1

⇒ tan² 2x = $\dfrac{1}{3}$

⇒ tan 2x = $\sqrt{\dfrac{1}{3}}$

⇒ tan 2x = $\dfrac{1}{\sqrt{3}}$

⇒ tan 2x = tan 30°

⇒ 2x = 30°

⇒ x = $\dfrac{30°}{2}$ = 15°.

Hence, x = 15°.

(ii) Given,

⇒ tan² x = 3(sec x - 1)

⇒ sec² x - 1 = 3sec x - 3

⇒ sec² x - 3 sec x - 1 + 3 = 0

⇒ sec² x - 3 sec x + 2 = 0

⇒ sec² x - 2 sec x - sec x + 2 = 0

⇒ sec x(sec x - 2) - 1(sec x - 2) = 0

⇒ (sec x - 1)(sec x - 2) = 0

⇒ sec x - 1 = 0 or sec x - 2 = 0

⇒ sec x = 1 or sec x = 2

⇒ sec x = sec 0° or sec x = sec 60°

⇒ x = 0° or x = 60°

Hence, x = 0° or 60°.

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Class	Frequencies
0-20	17
20-40	f₁
40-60	32
60-80	f₂
80-100	19

Mathematics

Solve for x ∈ W, 0° ≤ x ≤ 90°.
(i) 3 tan² 2x = 1
(ii) tan² x = 3(sec x - 1)

Trigonometric Identities

Answer

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Solve for x ∈ W, 0° ≤ x ≤ 90°.
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(ii) tan² x = 3(sec x - 1)

The mean of the following frequency distribution is 50, but the frequencies f₁ and f₂ in class 20-40 and 60-80 respectively are not known. Find these frequencies.
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0-20 17
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Given that the sum of frequencies is 120.

Prove that :
$\sqrt{\dfrac{\text{1+ sin A}}{\text{1 - sin A}}} - \sqrt{\dfrac{\text{1 - sin A}}{\text{1 + sin A}}}$ = 2 tan A

Prove that :
$\dfrac{\text{1 + cot A}}{\text{cos A}} + \dfrac{\text{1 + tan A}}{\text{sin A}}$ = 2(sec A + cosec A)