Question

Find the value the trigonometric terms: $\tan 67{\dfrac{1}{2}^ \circ } + \cot 67{\dfrac{1}{2}^ \circ }$ respectively.
A. $2\sqrt 2$
B. $2$
C. $\sqrt 2$
D. $1$

Answer 1

The given problem revolves around the concepts of trigonometric equations. So, we will use the definition of trigonometric equations and its identities. Here, we are going to extract the in bracket term i.e. angle then by considering the formula for trigonometric ratio for double angles say, $\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$ to find the given terms in an expression and then substituting the values the desired solution can be obtained.

Complete step by step answer:
Since, we have given the expression as,
$\tan 67{\dfrac{1}{2}^ \circ } + \cot 67{\dfrac{1}{2}^ \circ }$
As a result, the given expression can also be written as,
$\tan {\left( {\dfrac{{135}}{2}} \right)^ \circ } + \cot {\left( {\dfrac{{135}}{2}} \right)^ \circ }$
Where, $\dfrac{{135}}{2} = 67$
Now, let us assume that $\tan {\left( {\dfrac{{135}}{2}} \right)^ \circ } = \tan \theta$ for efficiency of the solution, we get
$\tan \theta + \cot \theta$
Since, we know that $\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
Hence,
$\tan 2\theta = \tan {135^ \circ } \\\ \Rightarrow \tan 2\theta = \tan ({90^ \circ } + {45^ \circ }) \\\$
According to the trigonometric conditions of change in four different quadrants (the above terminology exists in second quadrant), we get
$\tan 2\theta = - \cot {45^ \circ } \\\ \Rightarrow \tan 2\theta = - 1 \\\$
Where, $\cot {45^ \circ } = 1$
Now, hence considering the equation$\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$,
$\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = {\text{ }} - 1$
Solving the equation predominantly, we get

\Rightarrow 2\tan \theta = {\tan ^2}\theta - 1 \\\ \Rightarrow {\tan ^2}\theta - 2\tan \theta - 1 = 0 \\\ $$ As a result, the above equation seems to be quadratic in terms of $\tan \theta $. Let, $\tan \theta = x$. The equation becomes, $${x^2} - 2x - 1 = 0$$ The equation does not exists real factors, hence solving it by factorisation formula, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ Substituting the values in this equation, we get, $\Rightarrow x = \dfrac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4(1)( - 1)} }}{{2(1)}} \\\ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 + 4} }}{2} = \dfrac{{2 \pm \sqrt 8 }}{2} \\\ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 \times 2} }}{2} = \dfrac{{2 \pm 2\sqrt 2 }}{2} \\\ \Rightarrow x = \dfrac{{2(1 \pm \sqrt 2 )}}{2} \\\ \Rightarrow x = 1 \pm \sqrt 2 \\\ $ Re-substituting the value of$x = \tan \theta $, we get $$ \Rightarrow \tan \theta = 1 \pm \sqrt 2 $$ Considering the positive value of the equation (the angle never exists in negative standard), we get $$ \tan \theta = 1 + \sqrt 2 $$. Similarly, $$ \cot \theta = \dfrac{1}{{\tan \theta }} = \dfrac{1}{{1 + \sqrt 2 }}$$ Multiplying and dividing $1 - \sqrt 2 $ the above equation, we get $$\Rightarrow \cot \theta = \dfrac{1}{{1 + \sqrt 2 }} \times \dfrac{{1 - \sqrt 2 }}{{1 - \sqrt 2 }} \\\ \Rightarrow \cot \theta = \dfrac{{1 - \sqrt 2 }}{{{1^2} - {{\left( {\sqrt 2 } \right)}^2}}} \\\ \Rightarrow \cot \theta = \dfrac{{1 - \sqrt 2 }}{{1 - 2}} \\\ $$ Where, the algebraic identity $(a + b)(a - b) = {a^2} - {b^2}$ is been used, $$\Rightarrow \cot \theta = - (1 - \sqrt 2 ) \\\ \Rightarrow \cot \theta = \sqrt 2 - 1 \\\ $$ Now, hence the given equation becomes, i.e. $$\tan \theta + \cot \theta $$ Substituting the values that we have found in the above solution, we get $$\tan \theta + \cot \theta = 1 + \sqrt 2 + \sqrt 2 - 1 \\\ \Rightarrow \tan \theta + \cot \theta = 1 + \sqrt 2 + \sqrt 2 - 1 \\\ \Rightarrow \tan \theta + \cot \theta = 2\sqrt 2 \\\ $$ Re-Substitute the value of $\theta $assume earlier i.e. $\tan {\left( {\dfrac{{135}}{2}} \right)^ \circ } = \theta $, we get the required solution $ \therefore \tan {\left( {\dfrac{{135}}{2}} \right)^ \circ } + \cot {\left( {\dfrac{{135}}{2}} \right)^ \circ } = 2\sqrt 2 $ **Hence, option A is correct.** **Note:** One must know how to convert the ‘tan’, ‘cot’, ‘sec’ and ‘cosec’ terms in trigonometric formulae for all the terms especially trigonometric ratios for double angle, triple angles, half angles, compound angles, etc. say, $\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$ so as to distinguish the solution accurately. Also, we should know all the required values of standard angles say, $${0^o},{30^o},{45^o},{60^o},{90^o},{180^o},{270^o},{360^o}$$ respectively for each trigonometric term such as $\sin ,\cos ,\tan ,\cot ,\sec ,\cos ec$, etc. We should take care of the calculations so as to be sure of our final answer.