Question

# Find the value of $\dfrac{{\tan {{70}^0} - \tan {{20}^0}}}{{\tan {{50}^0}}}$ from the options given belowA. 2B. 1C. 0D. 3

Hint: Make use of the formula of tan(x-y) and express the numerator in a solvable form using this formula and solve it.

We have been asked to find out the value of $\dfrac{{\tan {{70}^0} - \tan {{20}^0}}}{{\tan {{50}^0}}}$
So, now we will make use of the formula
$\tan (x - y) = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}$
So, from this we get tanx-tany=tan(x-y)(1+tanxtany)
So, let us express the numerator in this from, so we get
$\dfrac{{\tan ({{70}^0} - {{20}^0})(1 + \tan {{70}^0}.\tan {{20}^0})}}{{\tan {{50}^0}}}$
Now, in the denominator,
$\tan {50^ \circ } = \tan ({70^ \circ } - {20^ \circ })$
So, we can write the expression as
$\dfrac{{\tan ({{70}^ \circ } - {{20}^ \circ })(1 + \tan {{70}^ \circ }.\tan {{20}^ \circ })}}{{\tan ({{70}^ \circ } - {{20}^ \circ })}}$
So, from this,
$\tan ({70^ \circ } - {20^ \circ })$ in the numerator and denominator would get cancelled out
So, we get the equation as:
$(1 + \tan {70^ \circ }\tan {20^ \circ })$
Now, this can be further written as $(1 + \tan ({90^ \circ } - {20^ \circ })\tan {20^ \circ })$
We , also know the result which says $\tan ({90^ \circ } - \theta ) = \cot \theta$
So, from this we get $1 + \cot {20^ \circ }\tan {20^ \circ }$
Since, cot and tan are reciprocal of each other,
We can write $1 + \dfrac{1}{{\tan {{20}^ \circ }}} \times \tan {20^ \circ }$
=1+1=2
So, from this, we can write $\dfrac{{\tan {{70}^0} - \tan {{20}^0}}}{{\tan {{50}^0}}} = 2$
So, option A is the correct answer for this question

Note: Make use of the appropriate trigonometric formula which is needed and then from that obtain the needed result , also make sure to mention the correct option after finding the solution, since it is a multiple choice question.