Given that x = secA + tanA, show that x + (1/x) = 2secA

[Monkeyseat]

Question:

Given that x = secA + tanA, show that x + (1/x) = 2secA

Working:

x = secA + tanA
x^2 = sec^2 A + tan^2 A
x^2 = sec^2 A + sec^2 A - 1
x^2 + 1 = 2sec^2 A

Square root both sides... (I don't know if you can do this)

x + 1 = 2secA

I don't think I've done that correctly. I think it might be alright up to x^2 + 1 = 2sec^2 A, but I don't know how to progress to x + 1/x = 2secA. Does anyone have any pointers?

Thanks for any help.

 

[soroban]

Hello, Monkeyseat!

\(\displaystyle \text{Given that: }\:x \:= \:\sec A + \tan A\text{, show that: }\:x + \frac{1}{x}\:=\: 2\sec A\)

\(\displaystyle \text{Substitute: }\;x + \frac{1}{x} \;=\;(\sec A + \tan A) + \frac{1}{\sec A + \tan A}\)


\(\displaystyle \text{Mulltiply the fraction by }\frac{\sec A - \tan A}{\sec A - \tan A}\)

. . \(\displaystyle (\sec A + \tan A) + \frac{1}{\sec A + \tan A}\cdot\frac{\sec A - \tan A}{\sec A - \tan A}\)

. . \(\displaystyle = \;(\sec A + \tan A) + \frac{\sec A - \tan A}{\underbrace{\sec^2\!A - \tan^2\!A}_{\text{This is 1}}}\)

. . \(\displaystyle = \;(\sec A + \tan A) + (\sec A - \tan A)\)

. . \(\displaystyle =\;2\sec A\)

 

[Deleted member 4993]

Question:

Given that x = secA + tanA, show that x + (1/x) = 2secA

Working:

x = secA + tanA
x^2 = sec^2 A + tan^2 A .........That is not correct
x^2 = sec^2 A + sec^2 A - 1
x^2 + 1 = 2sec^2 A

Square root both sides... (I don't know if you can do this)

x + 1 = 2secA

That is not correct because

\(\displaystyle (x \, + \, 1)^2 \, = x^2 \, + \, 1 \, + \, 2x\)

You really need to brush up your algebra


I don't think I've done that correctly. I think it might be alright up to x^2 + 1 = 2sec^2 A, but I don't know how to progress to x + 1/x = 2secA. Does anyone have any pointers?

Thanks for any help.

x = sec A + tan A

x^2 = sec^2 A + tan^2 A + 2 sec A * tan A = sec^2 + (sec^2 A - 1) + 2 sec A * tan A = 2sec^2 A + 2 sec A * tan A - 1

x + 1/x = (x^2 + 1)/x

Now continue.....

 

[Monkeyseat]

Ahhhh... I see where I've gone wrong Subhotosh Khan. Thanks both of you (again!).

Just one question soroban, why do you:

\(\displaystyle \text{Mulltiply the fraction by }\frac{\sec A - \tan A}{\sec A - \tan A}\) <--- Is that because this equals 1 anyway so does not alter the value of the fraction? Is that correct?

Do you multiply just to get the denominator as 1 (because 1 + tan^2 A = sec^2 A)?

Thanks.