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}
= {(b+a)/b}/√(1-(a²/b²))
= {(b+a)/b}/√{(b²-a²)/b²}
= {(b+a)/b}/√{(b²-a²)/b}
= (b+a)/√(b²-a²)
= √{(b+a)(b+a)/(b²-a²)}
⇛ LHS = √{(b+a)(b+a)/(b+a)(b-a)}
Now, (x+y)(x-y) = x²-y²
Where ,
On cancelling (b+a) then
⇛LHS = √{(b+a)/(b-a)}
⇛RHS
⇛ LHS = RHS
Sec θ + Tan θ = √{(b+a)/(b-a)}
Hence, Proved.
<u>Answer</u><u>:</u> If Sinθ=a/b then Secθ+Tanθ=√{(b+a)/(b-a)}.
<u>also</u><u> read</u><u> similar</u><u> questions</u><u>:</u><u>-</u> i) sin^2 A sec^2 B + tan^2 B cos^2 A = sin^2A + tan²B..
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Sec x -tan x sin x =1/secx Help me prove it..
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Answer:
5x+1
Step-by-step explanation:
Answer:
42 inches
Step-by-step explanation:
the top is adding 1 and the bottom is subtracting 5
Hi,
As of right now it is not likely for him to get a B. This is because his average is 63.6% so unless he gets better scores, he's stuck with an F.
