Question

${\tan }^3\theta +{\cot }^3\theta =12+8cose{c}^{3}2\theta$ if $\theta =$

• A. $\frac{7\pi }{12}$
• B. $\frac{11\pi }{12}$
• C. $\frac{19\pi }{12}$
• D. $\frac{23\pi }{12}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\frac{23\pi }{12}$
B $\frac{7\pi }{12}$
C $\frac{11\pi }{12}$
D $\frac{19\pi }{12}$

Convert $8{\text{cosec}}^3\left(2\theta \right)$ into $8\times \frac{1}{{\sin }^3\left(2\theta \right)}=8\times \frac{1}{(2\sin \theta \cos \theta {)}^3}=\frac{1}{{\sin }^3\theta {\cos }^3\theta }$

Similarly we write ${\tan }^3\theta$ and ${\cot }^3\theta$ in terms of $\sin \theta$ and $\cos \theta$.

Denote $\sin \theta =S$ and $\cos \theta =C$

So, $\frac{S^3}{C^3}+\frac{C^3}{S^3}=12+\frac{1}{S^3{C}^3}$

Take $L.C.M$, we get $\frac{S^6+C^6}{S^3{C}^3}=\frac{12S^3{C}^3+1}{S^3{C}^3}$

Now we cancel $S^3{C}^3$ but also note that means $S=0$ and $C=0$ can't be solution of the equation because they are in denominator.

$\Rightarrow \sin \theta \ne 0,\cos \theta \ne 0$

So equation now is, $S^6+C^6=12S^3{C}^3+1$

We know $S^6+C^6=1-3S^2{C}^2$.

Placing in above equation,

$1-3S^2{C}^2=12S^3{C}^3+1\to 3S^2{C}^2(4SC+1)=0$

$S\ne 0,C\ne 0\to 4SC+1=0\Rightarrow 4\sin \theta \cos \theta +1=0\Rightarrow 2\sin 2\theta +1=0$

$\sin 2\theta =-\frac{1}{2}\Rightarrow 2\theta =(2n+1)\pi +\frac{\pi }{6}$ and $2\theta =(2n+1)\pi +\frac{5\pi }{6}$

$\theta =(2n+1)\frac{\pi }{2}+\frac{\pi }{12}$ and $\theta =(2n+1)\frac{\pi }{2}+\frac{5\pi }{12}$

Hence put $n=0,1$ we get $\theta =\frac{7\pi }{12},\frac{11\pi }{12},\frac{19\pi }{12},\frac{23\pi }{12}$

Hence, $\left(A\right)\left(B\right)\left(C\right)\left(D\right)$