Question

secΘ + tanΘ = 3/2, Find sinΘ​

Answer 1

Question :- secΘ + tanΘ = 3/2, Find sinΘ ?

Solution :-

we know that,

• sec²A - tan²A = 1 .

So,

→ sec²Θ - tanΘ² = 1

Now, we also know that,

(a² - b²) = (a + b)(a - b) .

then,

→ (secΘ + tanΘ)(secΘ - tanΘ) = 1

→ (3/2)(secΘ - tanΘ) = 1

→ (secΘ - tanΘ) = (2/3) ------ Eqn.(1)

adding given , (secΘ + tanΘ) = 3/2 and Eqn.(1) , we get,

→ (secΘ + tanΘ) + (secΘ - tanΘ) = (3/2) + (2/3)

→ 2secΘ = (9 + 4)/6 = 13/6

→ secΘ = 13/12

→ secΘ = Hypotenuse/Base

comparing,

• Hypotenuse = 13
• Base = 12 .

Therefore,

→ Perpendicular = √[(Hypotenuse)² - (Base)²] { using pythagoras .}

→ Perpendicular = √[(13)² - (12)²] = √[169 - 144] = √(25) = 5 .

Hence,

→ sinΘ = Perpendicular/hypotenuse

→ sinΘ = 5/13 (Ans.)

Learn more :-

1) secΘ + tanΘ = 3/2, Find sinΘ

2) If 5sinΘ = 4, Find 5sinΘ - 3cosΘ/sinΘ + 2cosΘ

https://brainly.in/question/29119831

Answer 2

Answer:

Question ⬇

secΘ + tanΘ = 3/2, Find sinΘ

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we know that,

sec²A - tan²A = 1 .

So,

→ sec²Θ - tanΘ² = 1

Now, we also know that,

(a² - b²) = (a + b)(a - b) .

then,

→ (secΘ + tanΘ)(secΘ - tanΘ) = 1

→ (3/2)(secΘ - tanΘ) = 1

→ (secΘ - tanΘ) = (2/3) ------ Eqn.(1)

adding given , (secΘ + tanΘ) = 3/2 and Eqn.(1) , we get,

→ (secΘ + tanΘ) + (secΘ - tanΘ) = (3/2) + (2/3)

→ 2secΘ = (9 + 4)/6 = 13/6

→ secΘ = 13/12

→ secΘ = Hypotenuse/Base

Comparing,

Hypotenuse = 13

Base = 12 .

Therefore,

→ Perpendicular = √[(Hypotenuse)² - (Base)²] { using pythagoras .}

→ Perpendicular = √[(13)² - (12)²] = √[169 - 144] = √(25) = 5 .

Hence,

→ sinΘ = Perpendicular/hypotenuse

→ sinΘ = 5/13

Thanks..