Question

If sec \(\theta\) + tan \(\theta\) = m, show that \(\frac{m^2 - 1}{m^2 + 1} = \sin \theta\).

Hint:

Alternatively, you can apply componendo and dividendo on \(m^2 = \frac{1 + \sin \theta}{1 - \sin \theta}\):
\[ \frac{m^2 - 1}{m^2 + 1} = \frac{(1 + \sin \theta) - (1 - \sin \theta)}{(1 + \sin \theta) + (1 - \sin \theta)} = \frac{2 \sin \theta}{2} = \sin \theta \] This provides a highly structured and elegant way to skip fraction simplification!

Answer 1

Step 1: Find the reciprocal relation for sec and tan.
Since $(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=\sec^2\theta-\tan^2\theta=1$ and $\sec\theta+\tan\theta=m$, it follows that $\sec\theta-\tan\theta=\frac1m$.
Step 2: Solve the two equations together for sec and tan.
Adding: $2\sec\theta=m+\frac1m \implies \sec\theta=\frac{m^2+1}{2m}$. Subtracting: $2\tan\theta=m-\frac1m \implies \tan\theta=\frac{m^2-1}{2m}$.
Step 3: Divide tan by sec to get sin directly.
$\sin\theta=\frac{\tan\theta}{\sec\theta}=\frac{\frac{m^2-1}{2m}}{\frac{m^2+1}{2m}}=\frac{m^2-1}{m^2+1}$. Hence $\frac{m^2-1}{m^2+1}=\sin\theta$ is proved.