Question

The value of ${\tan ^3}\theta + {\cot ^3}\theta = 12 + 8\cos e{c^3}2\theta =$
A. $\dfrac{{7\pi }}{{12}}$
B. $\dfrac{{11\pi }}{{12}}$
C. $\dfrac{{19\pi }}{{12}}$
D. $\dfrac{{23\pi }}{{12}}$

Answer 1

Hint : As per the question we can see that it is related to trigonometry. Tangent, cot and cosecant are the trigonometric ratios. We will apply the trigonometric identities to solve this question. First we will convert the trigonometric ratios into their simpler forms i.e. in the terms of sine, cosine. We know that $\sin 2\theta = 2\sin \theta \cos \theta$ .

Complete step by step solution:
We know that $\cos ec\theta = \dfrac{1}{{\sin \theta }}$ , so we can write $8\cos e{c^3}2\theta = 8 \times \dfrac{1}{{{{\sin }^3}2\theta }}$ .
Now by applying the above formula in the denominator we can write $\dfrac{8}{{{{(2\sin \theta \cos\theta )}^3}}} = \dfrac{8}{{8{{\sin }^3}\theta {{\cos }^3}\theta }}$ . The $8$ will get cancelled out in the numerator and denominator i.e. $\dfrac{1}{{{{\sin }^3}\theta {{\cos }^3}\theta }}$ .
Similarly we can write ${\tan ^3}\theta = \dfrac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}$ and ${\cot ^3}\theta = \dfrac{{{{\cos }^3}\theta }}{{{{\sin }^3}\theta }}$ .
To solve the question in simpler form let us assume $\sin \theta = S$ and $\cos \theta = C$ .
Now let us write the given expression as
$\dfrac{{{S^3}}}{{{C^3}}} + \dfrac{{{C^3}}}{{{S^3}}} = 12 + \dfrac{1}{{{S^3}{C^3}}}$ .
By taking L.C.M we get
$\dfrac{{{S^6} + {C^6}}}{{{S^3}{C^3}}} = \dfrac{{12{S^3}{C^3} + 1}}{{{S^3}{C^3}}}$ , Now we cancel the denominator from both the sides of the equation, so it gives us
${S^6} + {C^6} = 12{S^3}{C^3} + 1$ .
We know that there is trigonometric identity which says that
${S^6} + {C^6} = 1 - 3{S^2}{C^2}$ .
Now by placing this in the above equation we have
$1 - 3{S^2}{C^2} = 12{S^3}{C^3} + 1$ .
We will now bring the similar terms together i.e. $1 - 1 = 12{S^3}{C^3} + 3{S^2}{C^2} \Rightarrow 0 = 12{S^3}{C^3} + 3{S^2}{C^2}$ .
By taking the common factor out we have $3{S^2}{C^2}(4SC + 1) = 0$ , we know that $S \ne 0,C \ne 0$ .
So we have $4SC + 1 = 0 \Rightarrow 4\sin \theta \cos \theta + 1 = 0$ .
We can write this equation as $2 \times 2\sin \theta \cos \theta = -1$ .
So by using the double angle formula we can write it as $2\sin 2\theta + 1 = 0$ .
Now we solve it: $\sin 2\theta = - \dfrac{1}{2}$ , by removing the sine we can write it as
$2\theta = (2n + 1)\pi + \dfrac{\pi }{6},(2n + 1)\pi + \dfrac{{5\pi }}{6}$ .
By isolating the term
$\theta$ , we have $\theta = \dfrac{1}{2}\left[ {(2n + 1)\pi + \dfrac{\pi }{6}} \right] = (2n + 1)\dfrac{\pi }{2} + \dfrac{\pi }{{12}}$ , similarly we isolating the tehta again
$\theta = \dfrac{1}{2}\left[ {(2n + 1)\pi + \dfrac{{5\pi }}{6}} \right] = (2n + 1)\dfrac{\pi }{2} + \dfrac{{5\pi }}{{12}}$ ,
We will put the value of $n = 0$ and $1$ , by each and calculate the value:
By putting zero and adding them, we have
$(2 \times 0 + 1)\dfrac{\pi }{2} + \dfrac{\pi }{{12}} = \dfrac{{6\pi + \pi }}{{12}} = \dfrac{{7\pi }}{{12}}$ ,
Again by putting zero on the other value
$(2 \times 0 + 1)\dfrac{\pi }{2} + \dfrac{{5\pi }}{{12}} = \dfrac{{6\pi + 5\pi }}{{12}} = \dfrac{{11\pi }}{{12}}$ .
We will now put the value of $n = 1$ , on the firm term:
$(2 \times 1 + 1)\dfrac{\pi }{2} + \dfrac{\pi }{{12}} = \dfrac{{18\pi + \pi }}{{12}} = \dfrac{{19\pi }}{{12}}$ ,
Similarly on the second value
$(2 \times 1 + 1)\dfrac{\pi }{2} + \dfrac{{5\pi }}{{12}} = \dfrac{{18\pi + 5\pi }}{{12}} = \dfrac{{23\pi }}{{12}}$ .
We can see that all four values of
$\theta = \dfrac{{7\pi }}{{12}},\dfrac{{11\pi }}{{12}},\dfrac{{19\pi }}{{12}},\dfrac{{23\pi }}{{12}}$ .
Hence the correct options are all i.e. (a),(b),(c),(d).
So, the correct answer is “Option A, B, C and D”.

Note : Before solving this kind of question we should be fully aware of the trigonometric formulas and their identities. One of the identities that we have used above is ${\sin ^6}\theta + {\cos ^6}\theta = 1 - 3{\sin ^2}\theta {\cos ^2}\theta$ . We should solve the question carefully to avoid any calculations mistakes.