Question

The value of
$\tan 7{\dfrac{1}{2}^ \circ } = \\\ A.\sqrt 6 - \sqrt 3 + \sqrt 2 - 2 \\\ B.\sqrt 6 - \sqrt 3 + \sqrt 2 - 1 \\\ C.\sqrt 5 - \sqrt 3 + \sqrt 2 - 2 \\\ D.\sqrt 6 - \sqrt 7 + \sqrt 2 - 2 \\\$

Answer 1

Hint : Consider the system of rectangular coordinate axes dividing the plane into four quadrants. These quadrants have different trends of sign for various trigonometric ratios. The quadrants are marked by the angle θ. The angle θ is said to be positive if measured in counter clockwise direction from the positive x-axis and is negative if measured in clockwise direction.
Formula used:
$\therefore \sin {45^ \circ } = \cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }} \\\ \therefore \cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2},\sin {30^ \circ } = \dfrac{1}{2} \;$

Complete step-by-step answer :
In the above question we are given with the angle of $7{\dfrac{1}{2}^ \circ }$ . We know that it will lie in the first quadrant as the first quadrant has all the angles less than 90°. In the first quadrant all the trigonometric ratios are positive.
Let us suppose the angle $7{\dfrac{1}{2}^ \circ }$ as angle A and substitute it at the end.
We know that tangent of A can be written as
$\tan A = \dfrac{{\sin A}}{{\cos A}}$
The above equation can also be written by multiplying 2sinA in both the numerator and denominator at the R.H.S of the equation such that they cancel each other,
$\tan A = \dfrac{{2{{\sin }^2}A}}{{2\cos A\sin A}}$
From the trigonometric ratios of multiples of an angle we know that
$\therefore \cos 2A = 1 - 2{\sin ^2}A \\\ \therefore \sin 2A = 2\sin A\cos A \;$
Therefore substituting the values from the formulas in the equation. We get,
$\tan A = \dfrac{{1 - \cos 2A}}{{\sin 2A}}$
Since we substituted the angle A therefore 2A will be
$A = 7{\dfrac{1}{2}^ \circ },2A = {15^ \circ }$
Therefore the above equation can be rewritten as
$\tan 7{\dfrac{1}{2}^ \circ }$
Now let us find the value of both $\cos {15^ \circ }$ and $\sin {15^ \circ }$ where
$\cos {15^ \circ } = \cos ({45^ \circ } - {30^ \circ }) \\\ \sin {15^ \circ } = \sin ({45^ \circ } - {30^ \circ }) \;$
Therefore using the property of compound angle made by sum of two or more angles,
Sin(A-B)=sinAcosB-cosAcosB
Cos(A-B)=cosAcosB+sinAsinB
Therefore values will be
$\cos {15^ \circ } = \cos {45^ \circ }\cos {30^ \circ } + \sin {45^ \circ }\sin {30^ \circ } \\\ = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2} = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \\\ \sin {15^ \circ } = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} \\\$
On substituting
$\tan 7{\dfrac{1}{2}^ \circ } = \dfrac{{1 - \cos {{15}^ \circ }}}{{\sin {{15}^ \circ }}} = \dfrac{{1 - \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}}}{{\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}}} = \dfrac{{2\sqrt 2 - \sqrt 3 - 1}}{{\sqrt 3 - 1}}$
Multiplying denominator and numerator by to remove the root in the denominator
$\dfrac{{2\sqrt 2 - \sqrt 3 - 1}}{{\sqrt 3 - 1}} \times \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}} = \sqrt 6 - \sqrt 3 + \sqrt 2 - 2$
So, the correct answer is “Option A”.

Note : All the six trigonometric functions have got a very important property in common that is periodicity. Remember that the trigonometric ratios are real numbers and remain the same as long as angle A is real.