Question

If tan θ + sec θ =e^x, then cos θ equals

Options

• \[\frac{e^x + e^{- x}}{2}\]

   

• \[\frac{2}{e^x + e^{- x}}\]

   

• \[\frac{e^x - e^{- x}}{2}\]

   

• \[\frac{e^x - e^{- x}}{e^x + e^{- x}}\]

   

Answer 1

\[\frac{2}{e^x + e^{- x}}\]

We have:
\[ \tan \theta + \sec \theta = e^x \]

\[ \sec \theta + \tan \theta = e^x \left( 1 \right)\]

\[ \Rightarrow \frac{1}{sec\theta + tan\theta} = \frac{1}{e^x}\]

\[ \Rightarrow \frac{\sec^2 \theta - \tan^2 \theta}{\sec \theta + \tan \theta} = \frac{1}{e^x}\]

\[ \Rightarrow \frac{\left( \sec \theta + \tan \theta \right)\left( \sec \theta - \tan \theta \right)}{\left( \sec \theta + \tan \theta \right)} = \frac{1}{e^x}\]

\[ \therefore sec\theta-\tan\theta = \frac{1}{e^x} \left( 2 \right)\]

Adding ( 1 ) and ( 2 ): 

\[2\sec \theta = e^x + \frac{1}{e^x}\]

\[ \Rightarrow 2\sec \theta = \frac{\left( e^x \right)^2 + 1}{e^x}\]

\[ \Rightarrow \sec \theta = \frac{e^{2x} + 1}{2 e^x}\]

\[ \Rightarrow \sec \theta = \frac{1}{2} \times \frac{e^{2x} + 1}{e^x}\]

\[ \Rightarrow \sec \theta = \frac{1}{2}\times\left( e^x + e^{- x} \right)\]

\[ \Rightarrow \frac{1}{\cos \theta} = \frac{e^x + e^{- x}}{2}\]

\[ \Rightarrow \cos\theta = \frac{2}{e^x + e^{- x}}\]