# If sec x = a + (1)/(4a) , then sec x + tan x = the options are, (a) a (b) 2a (c) 3a (d) 4a PLS HLP !!

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Rewrite `secx=a+1/(4a)` as `secx=(4a^2+1)/(4a)` (Combine `a+1/(4a)` )

Consider a right triangle. Using right triangle trigonometry definitions with an acute angle `x` we have `secx=("hypotenuse")/("adjacent")` where hypotenuse is the length of the hypotenuse and adjacent the length of the leg adjacent to the angle `x` .

Thus the hypotenuse of the triangle is `4a^2+1` and the side adjacent to angle `x` is `4a` . Then the side opposite angle `x` can be found using the pythagorean theorem:

`"opposite"^2+(4a)^2=(4a^2+1)^2`

`"opposite"^2+16a^2=16a^4+8a^2+1`

`"opposite"^2=16a^4-8a^2+1=(4a^2-1)^2`

`"opposite"=4a^2-1` (Using the positive root as length is nonnegative)

Again we use the right triangle definition for tangent: `tanx=("opposite")/("adjacent")` or `(4a^2-1)/(4a)`

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`secx+tanx=(4a^2+1)/(4a)+(4a^2-1)/(4a)=(8a^2)/(4a)=2a`

**so the answer is (b) 2A**

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