If \(\sec\theta=7x\) and \(\displaystyle\tan\theta=\frac{7}{x}\), what is the value of \(\displaystyle 98\left(x^2-\frac{1}{x^2}\right)+2\)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4
The identity \(\sec^2\theta-\tan^2\theta=1\) removes \(\theta\) and leaves an equation in \(x\) alone.
\(\displaystyle 49x^2-\frac{49}{x^2}=1\)
\(\displaystyle x^2-\frac{1}{x^2}=\frac{1}{49}\)
\(\displaystyle 98\times\frac{1}{49}+2=2+2=4\)