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Question: If we have a trigonometric expression \(\sec \theta +\tan \theta =x\) , then which of the following ...

If we have a trigonometric expression sec⁡θ+tan⁡θ=x\sec \theta +\tan \theta =x , then which of the following is the value of sec⁡θ\sec \theta .
A. x2+1x\dfrac{{{x}^{2}}+1}{x}
B. x2+12x\dfrac{{{x}^{2}}+1}{2x}
C. x2−12x\dfrac{{{x}^{2}}-1}{2x}
D. x2−1x\dfrac{{{x}^{2}}-1}{x}

Explanation

Solution

To find the value of sec⁡θ\sec \theta , we will consider sec⁡θ+tan⁡θ=x...(i)\sec \theta +\tan \theta =x...(i). Multiply and divide (i) with sec⁡θ−tan⁡θ\sec \theta -\tan \theta . Using sec⁡2θ−tan⁡2θ=1{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 and solving this further, we will get sec⁡θ−tan⁡θ=1x...(ii)\sec \theta -\tan \theta =\dfrac{1}{x}...(ii) . When we add equations (i) and (ii) and solve, the value of sec⁡θ\sec \theta can be obtained.

Complete step-by-step solution
It is given that sec⁡θ+tan⁡θ=x\sec \theta +\tan \theta =x . We have to find the value of sec⁡θ\sec \theta .
Let us consider sec⁡θ+tan⁡θ=x...(i)\sec \theta +\tan \theta =x...(i)
We have to multiply and divide equation (i) with sec⁡θ−tan⁡θ\sec \theta -\tan \theta . We will get
(sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)sec⁡θ−tan⁡θ=x\dfrac{\left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)}{\sec \theta -\tan \theta }=x
Now, let us expand the numerator. We know that a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) . Hence,
sec⁡2θ−tan⁡2θsec⁡θ−tan⁡θ=x\dfrac{{{\sec }^{2}}\theta -{{\tan }^{2}}\theta }{\sec \theta -\tan \theta }=x
We know that sec⁡2θ−tan⁡2θ=1{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 . Hence, the above equation becomes
1sec⁡θ−tan⁡θ=x\dfrac{1}{\sec \theta -\tan \theta }=x
This can be written as
sec⁡θ−tan⁡θ=1x...(ii)\sec \theta -\tan \theta =\dfrac{1}{x}...(ii)
Now, let us add equations (i) and (ii). This is shown below.
sec⁡θ+tan⁡θ+sec⁡θ−tan⁡θ=x+1x ⇒2sec⁡θ=x+1x \begin{aligned} & \sec \theta +\tan \theta +\sec \theta -\tan \theta =x+\dfrac{1}{x} \\\ & \Rightarrow 2\sec \theta =x+\dfrac{1}{x} \\\ \end{aligned}
When we solve the RHS, we will get
2sec⁡θ=x2+1x2\sec \theta =\dfrac{{{x}^{2}}+1}{x}
Now, we can find the value of sec⁡θ\sec \theta by taking 2 to the RHS. We will get
sec⁡θ=x2+12x\sec \theta =\dfrac{{{x}^{2}}+1}{2x}
Hence, the correct option is B.

Note: You must know the trigonometric identities to solve this question. You may make mistake when writing the identity sec⁡2θ−tan⁡2θ=1{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 as sec⁡2θ−tan⁡2θ=−1{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =-1 . We can also solve this question in an alternate method as shown below.
We know that if sec⁡θ+tan⁡θ=a\sec \theta +\tan \theta =a then sec⁡θ−tan⁡θ=1a\sec \theta -\tan \theta =\dfrac{1}{a} .
We have sec⁡θ+tan⁡θ=x...(i)\sec \theta +\tan \theta =x...(i) . Then,
sec⁡θ−tan⁡θ=1x...(ii)\sec \theta -\tan \theta =\dfrac{1}{x}...\left( ii \right)
Now, let us add (i) and (ii). We will get
sec⁡θ+tan⁡θ+sec⁡θ−tan⁡θ=x+1x\sec \theta +\tan \theta +\sec \theta -\tan \theta =x+\dfrac{1}{x}
⇒2sec⁡θ=x+1x\Rightarrow 2\sec \theta =x+\dfrac{1}{x}
When we solve the RHS, we will get
2sec⁡θ=x2+1x2\sec \theta =\dfrac{{{x}^{2}}+1}{x}
Now, we can find the value of sec⁡θ\sec \theta by taking 2 to the RHS. We will get
sec⁡θ=x2+12x\sec \theta =\dfrac{{{x}^{2}}+1}{2x}