Question

If sec A + tan A =x,then what is the value of sec A?​

Answer 1

Answer:

sec’ and ‘tan’ are the abbreviations used for the secant and tangent functions respectively. As functions, we need to know there arguments, i.e. what are we taking the secant and tangent of?

I’m going to assume that we are using the same argument for both functions; i’ll call it y . We thus have: sec(y)+tan(y)=x

Subtracting tan(y) from both sides of the equation, we have: sec(y)=x−tan(y)

Step-by-step explanation:

(sec(y)+tan(y))2=sec2(y)+2sec(y)tan(y)+tan2(y)=x2

Using the trigonometric identity tan2(y)=sec2(y)−1, we can rewrite this as:

2sec2(y)+2sec(y)tan(y)−1=x2

Adding 1−2sec2(y) to both sides:

2sec(y)tan(y)=(x2+1)−2sec2(y)

Squaring both sides:

4sec2(y)tan2(y)=(x2+1)2−4(x2+1)sec2(y)+4sec4(y)

Using the identity to replace the tan2(y):

4sec4(y)−4sec2(y)=(x2+1)2−4(x2+1)sec2(y)+4sec4(y)

Adding 4(x2+1)sec2(y)−4sec4(y) to both sides:

4(x2+1)sec2(y)−4sec2(y)=(x2+1)2

Using the identity again:

tan2(y)=sec2(y)−1=(x2+1)24x2−1=(x2+1)2−4x24x2=(x2−1)24x2⇒tan(y)=±x2−12x

Trying the four combinations for the values of sec(y) and tan(y) in our original equation, the only solution is:

sec(y)=x2+12x

tan(y)=x2−12x

[The other combinations sum to −x and ±1x ]

Answer: sec(y)=x2+12x