We are given the equation: $ \sec A + \tan A = p $
We will use the fundamental trigonometric identity: $ \sec^2 A - \tan^2 A = 1 $
Factor the identity as a difference of squares: $ (\sec A - \tan A)(\sec A + \tan A) = 1 $
Substitute the given value of $p$: $ (\sec A - \tan A)(p) = 1 $
Solve for $(\sec A - \tan A)$: $ \sec A - \tan A = \frac{1}{p} $
Now we have a system of two linear equations:
Add the two equations to eliminate $\tan A$: $ (\sec A + \tan A) + (\sec A - \tan A) = p + \frac{1}{p} $ $ 2 \sec A = p + \frac{1}{p} $
Combine the terms on the right side: $ 2 \sec A = \frac{p^2 + 1}{p} $
Isolate $\sec A$ by dividing by 2: $ \sec A = \frac{p^2 + 1}{2p} $
This matches option A.
The given equation can be reduced to
If sin2x = a - b√c, where a and b are natural numbers and c is prime number, then what is the value of a - b + 2c ?
Let θ be a positive angle. If the number of degrees in θ is divided by the number of radians in θ, then an irrational number 180 / π results. If the number of degrees in θ is multiplied by the number of radians in θ, then an irrational number 125π / 9 results. The angle θ must be equal to
What is sin 2α equal to?
If \(\sin θ = \frac{8}{{17}}\) , then find the value of tan θ.