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)}.
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Answer:
3.50
Step-by-step explanation:
each banana would cost 50 cent
Answer:
the answer is 59
Step-by-step explanation:
So to find the answer you have to know what the mixture is, what the trend is,
on day one he used 4 cans of blue and 6 cans of yellow
day two, 6 blue and 9 yellow
both are divisible by 2 and 3 respectively, additionally when you divide the number of blue cans Marc used by the number of yellow cans each day you get a sum of .66
those are the trends
so therefore day three would have the same trend.
Now on day three Marc will have 4 cans of blue paint and 5 cans of yellow paint remaining, what I did here to find the answer was divide numbers under 4 and 5 until I got a division problem that equaled .66
my answer was 2 blue cans and 3 yellow.
The answer is that on day three the highest number of cans that Marc can mix to get his favorite shade of green is 2 blue and 3 yellow