If $\sec \theta=m, \tan \theta=n$, then $\frac{1}{m}\left[m+n+\frac{1}{m+n}\right]=$

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Aicharya · Accepted Answer

Given, $\sec \theta=m, \tan \theta=n$ $\begin{aligned} \therefore \frac{1}{m} & \left(m+n+\frac{1}{m+n}\right) \ = & \frac{1}{\sec \theta}\left(\sec \theta+\tan \theta+\frac{1}{\sec \theta+\tan \theta}\right) \ \quad= & \frac{(\sec \theta+\tan \theta)^2+1}{\sec \theta(\sec \theta+\tan \theta)} \ = & \frac{\left\{(1+\sin \theta)^2+\cos ^2 \theta\right\} / \cos ^2 \theta}{(1+\sin \theta) / \cos ^2 \theta} \ \quad= & \frac{1+\sin ^2 \theta+2 \sin \theta+\cos ^2 \theta}{1+\sin \theta}=\frac{2(1+\sin \theta)}{1+\sin \theta} \ =2 \end{aligned}$

If \(\sec \theta=m, \tan \theta=n\), then \(\frac{1}{m}\left[m+n+\frac{1}{m+n}\right]=\)

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