If \(\sec\theta+\tan\theta=R\), then what is the value of \(\sec\theta-\tan\theta\)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{R}\)
The identity \(\sec^2\theta-\tan^2\theta=1\) factorises as a difference of squares, so the sum and the difference of \(\sec\theta\) and \(\tan\theta\) always multiply to 1.
\((\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=\sec^2\theta-\tan^2\theta=1\)
\(R\times(\sec\theta-\tan\theta)=1\)
\(\sec\theta-\tan\theta\displaystyle=\frac{1}{R}\)