Question

If \[\sec \theta + \tan \theta = k\], then \[\cos \theta \] equals
A) \[\dfrac{{\left( {{k^2} + 1} \right)}}{k}\]
B) \[\dfrac{{2k}}{{\left( {{k^2} + 1} \right)}}\]
C) \[\dfrac{k}{{\left( {{k^2} + 1} \right)}}\]
D) \[\dfrac{k}{{({k^2} - 1)}}\]

Answer 1

Hint
Here, we use the trigonometric formulation of \[\sec \theta \] and \[\tan \theta \] and after using it and make familiar with the given question we use algebraic identity and put in this identity. After making two equations, one is given in the question and the other that we have made by using identities and solving these two equations to get the result for \[\cos \theta \].

Formula used
Here we use two formulas that are shown below;
1. \[{\sec ^2}\theta - {\tan ^2}\theta = 1\]
2. \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]
3. \[\sec \theta = \dfrac{1}{{\cos \theta }}\]

Complete step-by-step solution
The given equation is
\[\sec \theta + \tan \theta = k\] ………(1)
As we know, the trigonometric identity is \[{\sec ^2}\theta - {\tan ^2}\theta = 1\].
Now, we will simplify the \[{\sec ^2}\theta - {\tan ^2}\theta = 1\] by using the formula \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\], we get
\[{\sec ^2}\theta - {\tan ^2}\theta = \left( {\sec \theta - \tan \theta } \right)\left( {\sec \theta + \tan \theta } \right)\]........ (2)
Further, we will substitute equation (2) in trigonometric identity, we get
\[\left( {\sec \theta - \tan \theta } \right)\left( {\sec \theta + \tan \theta } \right) = 1\]……..(3)\
Furthermore, we will substitute equation (1) into equation (3), we get
\[k(\sec \theta - \tan \theta ) = 1\]
Now, we will simplify above equation, we get
\[\sec \theta - \tan \theta = \dfrac{1}{k}\].......... (4)
Further, we will add the equation (1) and the equation (4) and cancel the terms \[\tan \theta \], we get;
\[\begin{array}{l}\sec \theta + \tan \theta + \sec \theta - \tan \theta = k + \dfrac{1}{k}\\2\sec \theta = k + \dfrac{1}{k}\end{array}\]
Now, we will simplify the above expression by taking the LCM, we get
\[\sec \theta = \dfrac{{{k^2} + 1}}{{2k}}\]
Furthermore, we will apply the formula \[\sec \theta = \dfrac{1}{{\cos \theta }}\], we get
\[\dfrac{1}{{\cos \theta }} = \dfrac{{{k^2} + 1}}{{2k}}\]
Now, we will cross multiply both sides, we get
\[\cos \theta = \dfrac{{2k}}{{{k^2} + 1}}\]

Hence, option B) is the correct answer.

Note
In this type of question, we should remember the basic properties and the relations of the trigonometric functions. Without these relations, we cannot get the correct answer. Another important formula for solving the question is the quadratic formula which is used to find the roots of the quadratic equation.