plz solve this question
Answers
Answer:
Given :-
Tan θ + Sin θ = a
Tan θ - Sin θ = β
To find :-
Show that a²-β² = 4√(aβ)
Solution :-
Given that :
Tan θ + Sin θ = a ------(1)
On squaring both sides then
=> (Tan θ + Sin θ)² = a²
=>Tan² θ + 2 Tan θ Sin θ + Sin² θ = a²--(2)
And
Tan θ - Sin θ = β ------(3)
On squaring both sides then
=> (Tan θ - Sin θ)² = β²
=>Tan² θ - 2 Tan θ Sin θ + Sin² θ =β²---(4)
On Subtracting (4) from (3)
=> (3)-(4)
=> (Tan² θ + 2 Tan θ Sin θ + Sin² θ) -
(Tan² θ - 2 Tan θ Sin θ + Sin² θ) = a²- β²
=>Tan² θ + 2 Tan θ Sin θ + Sin² θ -
Tan² θ + 2 Tan θ Sin θ - Sin² θ = a²- β²
=> 2 Tan θ Sin θ +2 Tan θ Sin θ = a²- β²
=>4 Tan θ Sin θ = a²- β² -----(5)
Now
a = Tan θ + Sin θ
β = Tan θ - Sin θ
a β = (Tan θ + Sin θ)(Tan θ - Sin θ)
=> a β = Tan² θ - Sin²θ
=> a β = ( Sin² θ/ Cos² θ) - Sin² θ
=> a β = Sin² θ[ (1/Cos² θ)-1]
=> a β = Sin² θ[(1- Cos² θ)/Cos² θ]
=> a β = Sin² θ × Sin² θ/ Cos² θ
=> a β = Sin² θ Tan² θ
=> a β = ( Sin θ Tan θ)²
=> Tan θ Sin θ = √(a β)-------(6)
On Substituting this value in (5)
From (5)&(6)
=> a²- β² = 4√(a β)
Hence, Proved.
Used formulae:-
Tan A = Sin A / Cos A
Sin² A + Cos² A = 1
(a+b)² = a²+2ab+b²
(a-b)² = a²-2ab+b²
Answer:
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