Question

4sin2theta-1=0
find theta and find cos2theta+tan2theta​

Answer 1

Given:

• Value of 4sin²∅-1=0.

To Find:

• The value of cos²∅+tan²∅.

Formulae Used:

• (a+b)(a-b)=a²-b².

Answer:

Taking the given equation,

⇒4sin²∅ -1=0.

⇒(2sin²∅) -(1)²=0.

⇒ (2sin∅+1)(2sin∅-1)=0.

⇒sin∅ = ½ , -½ .

We would we solving taking positive value.

⇒sin∅ = ½

⇒ sin∅ = sin30°.

⇒∅ = 30°.

Putting ∅ = 30° .

= cos²∅ +tan²∅ .

= cos²30°+tan²30°.

= (√3/2)²+(1/√3)².

= 3/4 + 1/3 .

= 4+9/13 .

= 13/13

= 1 .

Hence the required answer is 1.

Answer 2

Answer:

Given:

Value of 4sin²0-1=0.

To Find:

. The value of cos²Ø+tan²0.

Formulae Used:

(a+b)(a-b)=a²-b².

Taking the given equation,

→ 4sin²0 -1=0.

> (2sin²0) -(1)²=0.

→ (2sinØ+1)(2sin-1)=0.

→ sin Ø= 1/2, -1/2.

We would we solving taking positive value.

→ sinØ = 12

→ sinØ = sin30°.

Putting Ø = 30°.

= cos²Ø+tan²0.

= cos²30°+tan²30°.

=(√3/2)²+(1/√3)².

= 3/4 + 1/3.

= 4+9/13.

= 13/13

= 1.

Hence the required answer is 1.

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