Find the value of $\dfrac{{\tan {{70}^0} - \tan {{20}^0}}}{{\tan {{50}^0}}}$ from the options given below A. 2 B. 1 C. 0 D. 3
Answer
Verified
Hint: Make use of the formula of tan(x-y) and express the numerator in a solvable form using this formula and solve it.
Complete step-by-step answer: 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.
×
Sorry!, This page is not available for now to bookmark.