Answer:
Alternate Interior Angles
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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1+3x
Distribute -7 through the parenthesis
Answer:
Step-by-step explanation:
divide both side by 8
<u>Answer</u>:
![f^{-1}(x) = \sqrt[5]{\frac{x+7}{9} -10}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29%20%3D%20%5Csqrt%5B5%5D%7B%5Cfrac%7Bx%2B7%7D%7B9%7D%20-10%7D)
<u>Explanation</u>:
given : f(x) = 9( 
+ 10 ) - 7
replace x and y with each other.
- x = 9(
+ 10 ) - 7
solve for y
- x + 7 = 9( + 10 )
- =
- 10
- y =
![\sqrt[5]{\frac{x+7}{9} -10}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B%5Cfrac%7Bx%2B7%7D%7B9%7D%20-10%7D)
Therefore,
