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Answers
Explanation:
Given :-
Sin θ = a/b
To find :-
Prove that Sec θ + Tan θ = √[(b+a)/(b-a)] .
Solution :-
Method -1:-
Given that :-
Sin θ = a/b
On squaring both sides then
=> (Sin θ)² = (a/b)²
=> Sin² θ = a²/b²
On Subtracting above equation from 1 then
=> 1 - Sin² θ = 1-(a²/b²)
=> Cos² θ = (b²-a²)/b²
Since Sin² A + Cos² A = 1
=> 1/ Cos² θ = b²/(b²-a²)
=> Sec² θ = b²/(b²-a²)
=> Sec θ = √[b²/(b²-a²)]
=> Sec θ = b/√(b²-a²) ---------------(1)
We have
Sec² θ = b²/(b²-a²)
On subtracting 1 both sides then
=> Sec² θ - 1 = [b²/(b²-a²)] - 1
=> Tan² θ = [ {b²-(b²-a²)}/(b²-a²)]
Since Sec² A - Tan² A = 1
=> Tan² θ = [(b²-b²+a²)/(b²-a²)]
=>Tan² θ = a²/(b²-a²)
=> Tan θ = √[a²/(b²-a²)]
=> Tan θ = a/√(b²-a²) --------------(2)
On adding (1)&(2) then
Sec θ + Tan θ = [b/√(b²-a²)] + [a/√(b²-a²)]
=> Sec θ + Tan θ = [(b+a)/√(b²-a²)]
=> Sec θ + Tan θ = √[(b+a)(b+a)/(b²-a²)]
=> Sec θ + Tan θ = √[(b+a)(b+a)/(b+a)(b-a)]
Since (x+y)(x-y) = x²-y²
Where , x = b and y = a
On cancelling (b+a) then
=> Sec θ + Tan θ = √[(b+a)/(b-a)]
Hence, Proved.
Method -2:-
Given that
Sin θ = a/b
LHS = Sec θ + Tan θ
=>(1/Cos θ) + (Sin θ/ Cos θ)
=> (1+Sin θ)/Cos θ
We know that
Sin² A + Cos² A = 1
=> Cos² A = 1-Sin² A
=> Cos A =√(1-Sin² A)
LHS = (1+Sin θ)/√(1- Sin² θ)
=> LHS = [1+(a/b)]/√[1-(a/b)²]
=> LHS = [(b+a)/b]/√(1-(a²/b²))
=> LHS = [(b+a)/b]/√[(b²-a²)/b²]
=> LHS = [(b+a)/b]/√(b²-a²)/b]
=> LHS = (b+a)/√(b²-a²)
=> LHS = √[(b+a)(b+a)/(b²-a²)]
=> LHS = √[(b+a)(b+a)/(b+a)(b-a)]
Since (x+y)(x-y) = x²-y²
Where , x = b and y = a
On cancelling (b+a) then
=> LHS = √[(b+a)/(b-a)]
=> RHS
=> LHS = RHS
Sec θ + Tan θ = √[(b+a)/(b-a)]
Hence, Proved.
Answer:-
If Sinθ=a/b then Secθ+Tanθ=√[(b+a)/(b-a)]
Used Identities :-
- Sin² A + Cos² A = 1
- Sec² A - Tan² A = 1
Used formulae:-
- (x+y)(x-y) = x²-y²
- Sec A = 1/Cos A
- Tan A = Sin A/Cos A