Question

If \[\tan \theta + \sec \theta = {e^x}\], then the value of $\cos \theta $ equals:
(A) $\dfrac{{\left( {{e^x} + {e^{ - x}}} \right)}}{2}$
(B) $\dfrac{2}{{\left( {{e^x} + {e^{ - x}}} \right)}}$
(C) $\dfrac{{\left( {{e^x} - {e^{ - x}}} \right)}}{2}$
(D) $\dfrac{{\left( {{e^x} - {e^{ - x}}} \right)}}{{\left( {{e^x} + {e^{ - x}}} \right)}}$

Answer 1

Hint: The given problem requires us to simplify the given trigonometric expression. The question requires thorough knowledge of trigonometric functions, formulae and identities. We are given an equation involving the trigonometric functions $\sec \theta $ and $\tan \theta $ . We will first use the trigonometric ${\sec ^2}\theta - {\tan ^2}\theta = 1$ to find the value of trigonometric functions $\sec \theta $ and $\tan \theta $. Then, we find the value of $\cos \theta $ using some basic trigonometric formulae.

Complete answer: In the given question, we are given a trigonometric equation \[\tan \theta + \sec \theta = {e^x}\].
So, we have a trigonometric identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$.
Factorising the left side of the equation, we get,
$ \Rightarrow \left( {\sec \theta - \tan \theta } \right)\left( {\sec \theta + \tan \theta } \right) = 1$
Now, substituting the value of $\left( {\sec \theta + \tan \theta } \right)$, we get,
$ \Rightarrow \left( {\sec \theta - \tan \theta } \right)\left( {{e^x}} \right) = 1$
$ \Rightarrow \left( {\sec \theta - \tan \theta } \right) = \dfrac{1}{{{e^x}}}$
So, we have the value of $\left( {\sec \theta - \tan \theta } \right)$ as $\dfrac{1}{{{e^x}}}$.
Now, adding the equations $\left( 1 \right)$ and $\left( 2 \right)$, we get,
$ \Rightarrow \left( {\sec \theta - \tan \theta } \right) + \left( {\tan \theta + \sec \theta } \right) = \dfrac{1}{{{e^x}}} + {e^x}$
So, simplifying the equation by cancelling the like terms with opposite signs, we get,
$ \Rightarrow 2\sec \theta = {e^{ - x}} + {e^x}$
Now, we use the law of negative exponents, ${a^{ - x}} = \dfrac{1}{{{a^x}}}$.
Dividing the equation by $2$, we get,
$ \Rightarrow \sec \theta = \left( {\dfrac{{{e^{ - x}} + {e^x}}}{2}} \right)$
So, we get the value of secant of the angle as $\left( {\dfrac{{{e^{ - x}} + {e^x}}}{2}} \right)$ from the given equation. Now, we know that cosine and secant functions are reciprocal of each other. So, we get the value of cosine as,
$ \Rightarrow \cos \theta = \left( {\dfrac{2}{{{e^{ - x}} + {e^x}}}} \right)$
So, we get the value of cosine as $\left( {\dfrac{2}{{{e^{ - x}} + {e^x}}}} \right)$.
Therefore, option (B) is the correct answer.

Note:
The trigonometric identities are of vital importance for solving any question involving trigonometric functions and identities. All the trigonometric ratios can be converted into each other using the simple trigonometric identities. Such questions require thorough knowledge of trigonometric conversions and ratios. Algebraic operations and rules like transposition rules come into significant use while solving such problems.