Question

To determine if \(x = \ln (\sec \theta + \tan \theta )\), then \(\sec \theta = \cosh x\).

Short answer

If\(x = \ln (\sec \theta + \tan \theta )\), then \(\sec \theta = \cosh x\).

Step-by-step solution

Given information

The given value is, \(x = \ln (\sec \theta + \tan \theta )\).

Concept of identities

The identity used:\(1 + {\tan ^2}x = {\sec ^2}x\)

Use the identity and substitute the values for proof

The given value is, \(x = \ln (\sec \theta + \tan \theta )\).

Take natural logarithm on both sides.

\(\begin{aligned}{e^x} &= {e^{\ln (\sec \theta + \tan \theta )}}\\ &= \sec \theta + \tan \theta \\ &= \sec \theta + \sqrt {{{\sec }^2}\theta - 1} \\{\rm{Q}}1 + {\tan ^2}x &= {\sec ^2}x\end{aligned}\)

The value of \({e^{ - x}}\) is computed as shown below.

\(\begin{aligned}{e^{ - x}} = \frac{1}{{\sec \theta + \sqrt {{{\sec }^2}\theta - 1} }}\quad \left( {{e^{ - x}} = \frac{1}{{{e^x}}}} \right)\\ = \frac{{\sec x - \sqrt {{{\sec }^2}\theta - 1} }}{{\left( {\sec \theta + \sqrt {{{\sec }^2}\theta - 1} } \right)\left( {{{\sec }^{\left. {x - \sqrt {{{\sec }^2}\theta - 1} } \right)}}} \right.}}\\\quad = \frac{{\sec \theta - \sqrt {{{\sec }^2}\theta - 1} }}{{{{\sec }^2}\theta - \left( {{{\sec }^2}\theta - 1} \right)}}\\ = \sec \theta - \sqrt {{{\sec }^2}\theta - 1} \end{aligned}\)

Thus, the value of \({e^{ - x}} = \sec \theta - \sqrt {{{\sec }^2}\theta - 1} \).

Substitute \({e^x} = \sec \theta + \sqrt {{{\sec }^2}\theta - 1} \) and \({e^{ - x}} = \sec \theta - \sqrt {{{\sec }^2}\theta - 1} \) in \(\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}\).

\(\begin{aligned}\cosh x &= \frac{{\sec \theta + \sqrt {{{\sec }^2}\theta - 1} + \sec \theta - \sqrt {{{\sec }^2}\theta - 1} }}{2}\\ &= \frac{{2\sec \theta }}{2}\\ &= \sec \theta \end{aligned}\)

That is, \(\sec \theta = \cosh x\).

Hence, the required result is proved