If \(\sec\theta=\displaystyle\frac{13}{12}\), where \(0^\circ<\theta<90^\circ\), then what is the value of \(\displaystyle\frac{1-\tan^2\theta}{1+\tan^2\theta}\)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{119}{169}\)
The identity \(1+\tan^2\theta=\sec^2\theta\) gives the denominator at once and also gives \(\tan^2\theta\).
\(1+\tan^2\theta=\sec^2\theta=\displaystyle\frac{169}{144}\)
\(\tan^2\theta=\displaystyle\frac{169}{144}-1=\frac{25}{144}\), so \(1-\tan^2\theta=\displaystyle\frac{119}{144}\)
Value \(=\displaystyle\frac{119}{144}\div\frac{169}{144}=\frac{119}{169}\)