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Question: If \[\tan \theta + \sec \theta = {e^x}\], then the value of \(\cos \theta \) equals: (A) \(\dfrac...

If tanθ+secθ=ex\tan \theta + \sec \theta = {e^x}, then the value of cosθ\cos \theta equals:
(A) (ex+ex)2\dfrac{{\left( {{e^x} + {e^{ - x}}} \right)}}{2}
(B) 2(ex+ex)\dfrac{2}{{\left( {{e^x} + {e^{ - x}}} \right)}}
(C) (exex)2\dfrac{{\left( {{e^x} - {e^{ - x}}} \right)}}{2}
(D) (exex)(ex+ex)\dfrac{{\left( {{e^x} - {e^{ - x}}} \right)}}{{\left( {{e^x} + {e^{ - x}}} \right)}}

Explanation

Solution

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 secθ\sec \theta and tanθ\tan \theta . We will first use the trigonometric sec2θtan2θ=1{\sec ^2}\theta - {\tan ^2}\theta = 1 to find the value of trigonometric functions secθ\sec \theta and tanθ\tan \theta . Then, we find the value of cosθ\cos \theta using some basic trigonometric formulae.

Complete answer: In the given question, we are given a trigonometric equation tanθ+secθ=ex\tan \theta + \sec \theta = {e^x}.
So, we have a trigonometric identity sec2θtan2θ=1{\sec ^2}\theta - {\tan ^2}\theta = 1.
Factorising the left side of the equation, we get,
(secθtanθ)(secθ+tanθ)=1\Rightarrow \left( {\sec \theta - \tan \theta } \right)\left( {\sec \theta + \tan \theta } \right) = 1
Now, substituting the value of (secθ+tanθ)\left( {\sec \theta + \tan \theta } \right), we get,
(secθtanθ)(ex)=1\Rightarrow \left( {\sec \theta - \tan \theta } \right)\left( {{e^x}} \right) = 1
(secθtanθ)=1ex\Rightarrow \left( {\sec \theta - \tan \theta } \right) = \dfrac{1}{{{e^x}}}
So, we have the value of (secθtanθ)\left( {\sec \theta - \tan \theta } \right) as 1ex\dfrac{1}{{{e^x}}}.
Now, adding the equations (1)\left( 1 \right) and (2)\left( 2 \right), we get,
(secθtanθ)+(tanθ+secθ)=1ex+ex\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,
2secθ=ex+ex\Rightarrow 2\sec \theta = {e^{ - x}} + {e^x}
Now, we use the law of negative exponents, ax=1ax{a^{ - x}} = \dfrac{1}{{{a^x}}}.
Dividing the equation by 22, we get,
secθ=(ex+ex2)\Rightarrow \sec \theta = \left( {\dfrac{{{e^{ - x}} + {e^x}}}{2}} \right)
So, we get the value of secant of the angle as (ex+ex2)\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,
cosθ=(2ex+ex)\Rightarrow \cos \theta = \left( {\dfrac{2}{{{e^{ - x}} + {e^x}}}} \right)
So, we get the value of cosine as (2ex+ex)\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.