Question

If $\tan \theta = a - \dfrac{1}{{4a}}$ , then $\sec \theta - \tan \theta$ equals to _________
A. $2a$
B. $\dfrac{1}{{2a}},-2a$
C. $- \dfrac{1}{{2a}},2a$
D. $- a,\dfrac{1}{{2a}}$

Answer 1

Hint : We use the identity $1 + {\tan ^2}\theta = {\sec ^2}\theta$ to solve the given problem. By using this identity we can find the value of $\sec \theta$ and substituting this in the given problem we get the required result. We also know $({a^2} + {b^2}) = {a^2} + {b^2} + 2ab$ and we are going to use this formula while doing simplification.

Complete step-by-step answer :
Let assume that $k = \sec \theta - \tan \theta {\text{ - - - - - - (1)}}$ . Now we need to find the value of ‘k’.
Now we have the identity $1 + {\tan ^2}\theta = {\sec ^2}\theta$ ,
Rearranging we have,
$\Rightarrow {\sec ^2}\theta = 1 + {\tan ^2}\theta$
Taking square root on both side we will have,
$\Rightarrow \sec \theta = \sqrt {1 + {{\tan }^2}\theta } {\text{ - - - - - (2)}}$
Now substituting equation (2) in equation (1) we have,
$k = \sqrt {1 + {{\tan }^2}\theta } - \tan \theta$
But in the given problem we have, $\tan \theta = a - \dfrac{1}{{4a}}$ .
Substituting in ‘k’ we have,
$\Rightarrow \sqrt {1 + {{\left( {a - \dfrac{1}{{4a}}} \right)}^2}} - \left( {a - \dfrac{1}{{4a}}} \right)$
We know $({a^2} - {b^2}) = {a^2} + {b^2} - 2ab$ , applying in above we have,
$\Rightarrow \sqrt {1 + {a^2} + {{\left( {\dfrac{1}{{4a}}} \right)}^2} - 2.a.\dfrac{1}{{4a}}} - a + \dfrac{1}{{4a}}$
Cancelling the terms we have,
$\Rightarrow \sqrt {1 + {a^2} + {{\left( {\dfrac{1}{{4a}}} \right)}^2} - \dfrac{1}{2}} - a + \dfrac{1}{{4a}}$
$\Rightarrow \sqrt {1 - \dfrac{1}{2} + {a^2} + {{\left( {\dfrac{1}{{4a}}} \right)}^2}} - a + \dfrac{1}{{4a}}$
We know $1 - \dfrac{1}{2} = \dfrac{1}{2}$ , then
$\Rightarrow \sqrt {\dfrac{1}{2} + {a^2} + {{\left( {\dfrac{1}{{4a}}} \right)}^2}} - a + \dfrac{1}{{4a}}$
We know that ${\left( {a + \dfrac{1}{{4a}}} \right)^2} = {a^2} + {\left( {\dfrac{1}{{4a}}} \right)^2} + 2.a.\dfrac{1}{{4a}} = {a^2} + {\left( {\dfrac{1}{{4a}}} \right)^2} + \dfrac{1}{2}$
Replacing the terms we have,
$\Rightarrow \sqrt {{{\left( {a + \dfrac{1}{{4a}}} \right)}^2}} - a + \dfrac{1}{{4a}}$
After cancelling the square and square root we will have,
$\Rightarrow \pm \left( {a + \dfrac{1}{{4a}}} \right) - a + \dfrac{1}{{4a}}$
That is two values,
Now taking positive sign $\Rightarrow + \left( {a + \dfrac{1}{{4a}}} \right) - a + \dfrac{1}{{4a}}$
$\Rightarrow a + \dfrac{1}{{4a}} - a + \dfrac{1}{{4a}}$
Cancelling ‘a’ we get,
$\Rightarrow \dfrac{1}{{4a}} + \dfrac{1}{{4a}}$
Taking L.C.M we have,
$\Rightarrow \dfrac{{1 + 1}}{{4a}}$
$\Rightarrow \dfrac{2}{{4a}}$
Cancelling we have
$\Rightarrow \dfrac{1}{{2a}}$
Now taking negative sign value $\Rightarrow - \left( {a + \dfrac{1}{{4a}}} \right) - a + \dfrac{1}{{4a}}$
$\Rightarrow - a - \dfrac{1}{{4a}} - a + \dfrac{1}{{4a}}$
Cancelling we have,
$\Rightarrow - a - a$
$\Rightarrow - 2a$ .
Thus we have $k = \dfrac{1}{{2a}},-2a$ .
So, the correct answer is “Option B”.

Note : Know the difference between ${\tan ^2}\theta$ and $\tan 2\theta$ . Both are different. That is, we know ${\tan ^2}\theta = {\left( {\tan \theta } \right)^2}$ and $\tan 2\theta$ is two multiplied with the angle. Don’t get confused with these two. Remember the important and identities formula in trigonometric functions. In above instead of $\sec \theta$ if we have $\cot \theta$ we would have solved this easily. Because tangent reciprocal is cotangent. By directly substituting we would get the answer.