#Quality Question
@Trigonometry
If θ is Acute angle and 3sinθ = 4cosθ then Find out the value of 4sin2θ - 3cos2θ + 2.
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Answered by
180
༒ Question ➽
If θ is Acute angle and 3sinθ = 4cosθ then Find out the value of 4sin2θ - 3cos2θ + 2.
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༒ Given ➽
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༒ To Find ➽
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༒ Solution ➽
As
So,
Hence,
So,
As θ is Acute so sinθ will not negative
Hence,
Now,
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༒ Alernate Solution ➽
As
Squaring We get
We Know
Now,
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Answered by
83
Given :-
If θ is angle and 3sinθ = 4cosθ
To Find :-
the value of 4sin²θ - 3cos²θ + 2.
Solution :-
3 sinθ = 4 cosθ
4/3 = sinθ/cosθ
Now, We know that
sinθ/cosθ = tanθ
4/3 = tanθ
On squaring both sides we get
(4/3)² = (tanθ)²
16/9 = tan²θ
sec²θ = 1 + tan²θ
sec²θ = 1 + 16/9
sec²θ = 9 + 16/9
sec²θ = 25/9
On rooting both sides
√(sin²θ) = √(25/9)
sinθ = 5/3
Now
cosθ = 1/sinθ
cosθ = 1/(5/3)
cosθ = 3/5
Now
sinθ = √(1 - cos²θ)
sinθ = √[1 - (3/5)²]
sinθ = √[1 - 9/25]
sinθ = √[25 - 9/25]
sinθ = √[16/25]
sinθ = 4/5
Finding the value
4 × (4/5)² - 3 × (3/5)² + 2
4 × (16/25) - 3 × (9/25) + 2
64/25 - 27/25 + 2
87/25
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