Question

1) secΘ + tanΘ = 3/2, Find sinΘ
2) If 5sinΘ = 4, Find 5sinΘ - 3cosΘ/sinΘ + 2cosΘ​

Answer 1

Question 1) :- 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.)

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

Solution :-

→ 5sinΘ = 4

→ sinΘ = 4/5 = Perpendicular / Hypotenuse .

we get,

• Perpendicular = 4.
• Hypotenuse = 5.

then,

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

→ Base = √[(5)² - (4)²] = √[25 - 16] = √(9) = 3 .

then,

• cosΘ = Base / Hypotenuse = 3/5 .

Putting all values now we get,

→ (5sinΘ - 3cosΘ)/(sinΘ + 2cosΘ)

→ (5*4/5 - 3*3/5) / (4/5 + 2*3/5)

→ (20/5 - 9/5) / (4/5 + 6/5)

→ (11/5) / (10/5)

→ (11/5) * (5/10)

→ (11/10) (Ans.)

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Answer 2

Answer:

1.sinθ and tanθ are negative, so θ lies in fourth quadrant.

2.Refers to above attachment.....