secΘ + tanΘ = 3/2, Find sinΘ
Answers
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
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Answer:
Question ⬇
secΘ + tanΘ = 3/2, Find sinΘ
⬇
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