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.