If \(\sec\theta=\displaystyle\frac{5}{4}\), 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 3
The correct answer is Option (3) → \(\displaystyle\frac{7}{25}\)
The identity \(1+\tan^2\theta=\sec^2\theta\) gives the denominator at once and also gives \(\tan^2\theta\).
\(1+\tan^2\theta=\sec^2\theta=\displaystyle\frac{25}{16}\)
\(\tan^2\theta=\displaystyle\frac{25}{16}-1=\frac{9}{16}\), so \(1-\tan^2\theta=\displaystyle\frac{7}{16}\)
Value \(=\displaystyle\frac{7}{16}\div\frac{25}{16}=\frac{7}{25}\)