Question

For \[\alpha =\dfrac{\pi }{7}\] which of the following hold(s) good?
(This question has multiple correct options)
\[\left( \text{a} \right)\text{ }\tan \alpha \tan 2\alpha \tan 3\alpha =\tan 3\alpha -\tan 2\alpha -\tan \alpha \]
\[\left( \text{b} \right)\text{ cosec}\alpha =\text{cosec}2\alpha +\text{cosec}4\alpha \]
\[\left( \text{c} \right)\text{ cos}\alpha -\cos 2\alpha +\cos 3\alpha =\dfrac{1}{2}\]
\[\left( \text{d} \right)\text{ 8cos}\alpha \cos 2\alpha \cos 4\alpha =1\]

Answer 1

Hint: To solve the given question, we will check each option one by one. For checking option (a), we will take \[\alpha +2\alpha =3\alpha .\] We will take tan on both sides. Then we will simplify it and match with the option. For checking the option (b), we will write the terms on LHS in the form of the sine function. Then we will simplify this by using the necessary trigonometric identities and putting the value of \[\alpha =\dfrac{\pi }{7}.\] For checking option (c), we will write \[\cos \alpha +\cos 3\alpha \] in terms of the product. Then we will multiply \[\sin \alpha \] on the numerator and denominator and simplify it. Finally, we will put the value of \[\alpha =\dfrac{\pi }{7}\] and check its correctness. For checking option (d), we will do the same steps as in the option (c).

Complete step-by-step answer:
It is given in the question that \[\alpha =\dfrac{\pi }{7}\] and we have to check which of the option will be correct when we will put \[\alpha =\dfrac{\pi }{7}\] on them. So, we will check each option one by one.
\[\text{Option}\left( \text{a} \right)\text{: }\tan \alpha \tan 2\alpha \tan 3\alpha =\tan 3\alpha -\tan 2\alpha -\tan \alpha \]
For checking this option, we have, \[\alpha +2\alpha =3\alpha .\]
Now, we will take tan on both the sides, so, we will get,
\[\Rightarrow \tan \left( a+2\alpha \right)=\tan 3\alpha .....\left( i \right)\]
Now, we will apply the identity given below
\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]
On applying this identity in LHS of (i), we will get,
\[\Rightarrow \dfrac{\tan \alpha +\tan 2\alpha }{1-\tan \alpha \tan 2\alpha }=\tan 3\alpha \]
On cross multiplying, we will get,
\[\Rightarrow \tan \alpha +\tan 2\alpha =\tan 3\alpha \left( 1-\tan \alpha \tan 2\alpha \right)\]
\[\Rightarrow \tan \alpha +\tan 2\alpha =\tan 3\alpha -\tan \alpha \tan 2\alpha \tan 3\alpha \]
\[\Rightarrow \tan \alpha \tan 2\alpha \tan 3\alpha =\tan 3\alpha -\tan \alpha -\tan 2\alpha \]
\[\text{Option}\left( \text{b} \right)\text{: cosec}\alpha =\text{cosec}2\alpha +\text{cosec}4\alpha \]
For checking this option, we will consider the right-hand side of the above equation. Thus, we have,
\[RHS=\text{cosec}2\alpha +\text{cosec}4\alpha \]
Now, we will convert \[\operatorname{cosec}\theta \] to sine form by the formula \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }.\]
\[\Rightarrow RHS=\dfrac{1}{\sin 2\alpha }+\dfrac{1}{\sin 4\alpha }\]
\[\Rightarrow RHS=\dfrac{\sin 4\alpha +\sin 2\alpha }{\sin 2\alpha \sin 4\alpha }\]
Now, we will apply the following trigonometric identities. In the numerator, we will apply,
\[\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\]
In the denominator, we will apply,
\[\sin 2\theta =2\sin \theta \cos \theta \]
Thus, we will get,
\[\Rightarrow RHS=\dfrac{2\sin \left( \dfrac{4\alpha +2\alpha }{2} \right)\cos \left( \dfrac{4\alpha -2\alpha }{2} \right)}{2\sin \alpha \cos \alpha \sin 4\alpha }\]
\[\Rightarrow RHS=\dfrac{2\sin 3\alpha \cos \alpha }{2\sin \alpha \cos \alpha \sin 4\alpha }\]
\[\Rightarrow RHS=\dfrac{\sin 3\alpha }{\sin \alpha \sin 4\alpha }\]
Now, we will put the value \[\alpha =\dfrac{\pi }{7}\] in the above equations.
\[\Rightarrow RHS=\dfrac{\sin \left( \dfrac{3\pi }{7} \right)}{\sin \left( \dfrac{\pi }{7} \right)\sin \left( \dfrac{4\pi }{7} \right)}\]
\[\Rightarrow RHS=\dfrac{\sin \left( \pi -\dfrac{4\pi }{7} \right)}{\sin \left( \dfrac{\pi }{7} \right)\sin \left( \dfrac{4\pi }{7} \right)}\]
Now, we can write \[\sin \left( \pi -\theta \right)=\sin \theta .\] So, we will get,
\[\Rightarrow RHS=\dfrac{\sin \left( \dfrac{4\pi }{7} \right)}{\sin \left( \dfrac{\pi }{7} \right)\sin \left( \dfrac{4\pi }{7} \right)}\]
\[\Rightarrow RHS=\dfrac{1}{\sin \left( \dfrac{\pi }{7} \right)}\]
\[\Rightarrow RHS=\operatorname{cosec}\left( \dfrac{\pi }{7} \right)\]
\[\Rightarrow RHS=\operatorname{cosec}\alpha \]
Hence LHS = RHS.
\[\text{Option}\left( \text{c} \right)\text{: cos}\alpha -\cos 2\alpha +\cos 3\alpha =\dfrac{1}{2}\]
 For checking this option, we will consider the LHS of the above equation. Thus, we will get,
\[LHS=\text{cos}\alpha -\cos 2\alpha +\cos 3\alpha \]
\[\Rightarrow LHS=\left( \text{cos}\alpha +\cos 3\alpha \right)-\cos 2\alpha \]
Now, we will apply the identity,
\[\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\]
Thus, we will get,
\[\Rightarrow LHS=2\cos \left( \dfrac{\alpha +3\alpha }{2} \right)\cos \left( \dfrac{3\alpha -\alpha }{2} \right)-\cos 2\alpha \]
\[\Rightarrow LHS=2\cos 2\alpha \cos \alpha -\cos 2\alpha \]
\[\Rightarrow LHS=\cos 2\alpha \left( 2\cos \alpha -1 \right)\]
Now, we will multiply both the numerator and denominator by \[\sin \alpha .\] Thus, we will get,
\[\Rightarrow LHS=\dfrac{\cos 2\alpha \left( 2\cos \alpha -1 \right)\sin \alpha }{\sin \alpha }\]
\[\Rightarrow LHS=\dfrac{\cos 2\alpha \left( 2\cos \alpha \sin \alpha -\sin \alpha \right)}{\sin \alpha }\]
Now, we know that \[\sin 2\theta =2\sin \theta \cos \theta .\] Thus, we will get,
\[\Rightarrow LHS=\dfrac{\cos 2\alpha \left( \sin 2\alpha -\sin \alpha \right)}{2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}\]
Now, we will use the identity
\[\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)\]
Thus, we will get,
\[\Rightarrow LHS=\dfrac{\cos 2\alpha \left[ 2\cos \left( \dfrac{2\alpha +\alpha }{2} \right)\sin \left( \dfrac{2\alpha -\alpha }{2} \right) \right]}{2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}\]
\[\Rightarrow LHS=\dfrac{\cos 2\alpha \left[ 2\cos \dfrac{3\alpha }{2}\sin \dfrac{\alpha }{2} \right]}{2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}\]
\[\Rightarrow LHS=\dfrac{\cos 2\alpha \cos \dfrac{3\alpha }{2}}{\cos \dfrac{\alpha }{2}}\]
Now, we will put \[\alpha =\dfrac{\pi }{7}\] in the above equation. Thus, we will get,
\[\Rightarrow LHS=\dfrac{\cos \dfrac{2\pi }{7}\times \cos \dfrac{3\pi }{14}}{\cos \dfrac{\pi }{14}}\]
Multiplying 2 on both the numerator and denominator, we get,
\[\Rightarrow LHS=\dfrac{2\cos \dfrac{2\pi }{7}\times \cos \dfrac{3\pi }{14}}{2\cos \dfrac{\pi }{14}}\]
\[\Rightarrow LHS=\dfrac{2\cos \left( \dfrac{\pi }{2}-\dfrac{3\pi }{14} \right)\cos \dfrac{3\pi }{14}}{2\cos \dfrac{\pi }{14}}\]
We know that,
\[\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta \]
So, we will get,
\[\Rightarrow LHS=\dfrac{2\sin \dfrac{3\pi }{14}\cos \dfrac{3\pi }{14}}{2\cos \dfrac{\pi }{14}}\]
We know that, \[\sin 2\theta =2\sin \theta \cos \theta .\] So, we get,
\[\Rightarrow LHS=\dfrac{\sin \dfrac{6\pi }{14}}{2\cos \dfrac{\pi }{14}}\]
\[\Rightarrow LHS=\dfrac{\sin \dfrac{6\pi }{14}}{2\cos \left( \dfrac{\pi }{2}-\dfrac{6\pi }{14} \right)}\]
On using the identity, \[\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta ,\] we will get,
\[\Rightarrow LHS=\dfrac{\sin \dfrac{6\pi }{14}}{2\sin \dfrac{6\pi }{14}}\]
\[\Rightarrow LHS=\dfrac{1}{2}\]
Hence, LHS = RHS.
\[\text{Option}\left( \text{d} \right)\text{: 8cos}\alpha \cos 2\alpha \cos 4\alpha =1\]
For checking this option, we will consider the LHS of the above equation. Thus, we will get,
\[LHS=\text{8cos}\alpha \cos 2\alpha \cos 4\alpha \]
Now, we will multiply the numerator and denominator by \[\sin \alpha ,\] so we will get,
\[\Rightarrow LHS=\dfrac{\text{8cos}\alpha \cos 2\alpha \cos 4\alpha \sin \alpha }{\sin \alpha }\]
\[\Rightarrow LHS=\dfrac{\text{4}\left( \text{2sin}\alpha \text{cos}\alpha \right)\cos 2\alpha \cos 4\alpha }{\sin \alpha }\]
On using sin 2A = 2 sin A cos A, we get,
\[\Rightarrow LHS=\dfrac{\text{4sin2}\alpha \cos 2\alpha \cos 4\alpha }{\sin \alpha }\]
\[\Rightarrow LHS=\dfrac{2\left( 2\sin 2\alpha \cos 2\alpha \right)\cos 4\alpha }{\sin \alpha }\]
\[\Rightarrow LHS=\dfrac{2\sin 4\alpha \cos 4\alpha }{\sin \alpha }\]
\[\Rightarrow LHS=\dfrac{\sin 8\alpha }{\sin \alpha }\]
Now, we will put \[\alpha =\dfrac{\pi }{7}\] in the above equation. Thus, we will get,
\[\Rightarrow LHS=\dfrac{\sin \left( \dfrac{8\pi }{7} \right)}{\sin \dfrac{\pi }{7}}\]
\[\Rightarrow LHS=\dfrac{\sin \left( \pi +\dfrac{\pi }{7} \right)}{\sin \dfrac{\pi }{7}}\]
We know that, \[\sin \left( \pi +\theta \right)=-\sin \theta ,\] so we get,
\[\Rightarrow LHS=\dfrac{-\sin \left( \dfrac{\pi }{7} \right)}{\sin \dfrac{\pi }{7}}\]
\[\Rightarrow LHS=-1\]
\[\Rightarrow LHS\ne RHS\]
Hence, options (a), (b) and (c) are the right options.

Note:The option (a) can also be checked in the following way.
\[\alpha +2\alpha +4\alpha =7a\]
We know that, \[\alpha =\dfrac{\pi }{7},\] so,
\[\Rightarrow \alpha +2\alpha +4\alpha =7\left( \dfrac{\pi }{7} \right)\]
\[\Rightarrow \alpha +2\alpha +4\alpha =\pi \]
Now, we know that if \[A+B+C=\pi \] then we have the following relation.
\[\tan A+\tan B+\tan C=\tan A\tan B\tan C\]
Thus, we have,
\[\tan \alpha +\tan 2\alpha +\tan 4\alpha =\tan \alpha \tan 2\alpha \tan 4\alpha \]
Now, we know that,
\[\tan 4\alpha =\tan \dfrac{4\pi }{7}=\tan \left( \pi -\dfrac{3\pi }{7} \right)\]
\[\tan \left( \pi -\theta \right)=-\tan \theta \]
Therefore,
\[\tan 4\alpha =-\tan \dfrac{3\pi }{7}=-\tan 3\alpha \]
Thus, we will get,
\[\tan \alpha +\tan 2\alpha -\tan 3\alpha =-\tan \alpha \tan 2\alpha \tan 3\alpha \]
\[\Rightarrow \tan 3\alpha -\tan \alpha -\tan 2\alpha =\tan \alpha \tan 2\alpha \tan 3\alpha \]
The option (c) can also be checked in the following way
\[LHS=8\cos \alpha \cos 2\alpha \cos 4\alpha \]
\[\Rightarrow LHS=8\cos \left( {{2}^{0}}\alpha \right)\cos \left( {{2}^{1}}\alpha \right)\cos \left( {{2}^{2}}\alpha \right)\]
Now, we know that if A is an acute angle, then we have the following relation
\[\cos A\times \cos 2A\times \cos {{2}^{2}}A.....\cos {{2}^{n-1}}A=\dfrac{\sin {{2}^{n}}A}{{{2}^{n}}\sin A}\]
In our case, n = 3, so we will get,
\[\Rightarrow LHS=\dfrac{8\times \sin \left( {{2}^{3}}\alpha \right)}{{{2}^{3}}\sin \alpha }\]
\[\Rightarrow LHS=\dfrac{8\sin 8\alpha }{8\sin \alpha }\]
\[\Rightarrow LHS=\dfrac{\sin 8\alpha }{\sin \alpha }\]
Now, \[\alpha =\dfrac{\pi }{7},\] so we get,
\[\Rightarrow LHS=\dfrac{\sin \dfrac{8\pi }{7}}{\sin \dfrac{\pi }{7}}\]
\[\Rightarrow LHS=\dfrac{\sin \left( \pi +\dfrac{\pi }{7} \right)}{\sin \dfrac{\pi }{7}}\]
\[\Rightarrow LHS=\dfrac{-\sin \dfrac{\pi }{7}}{\sin \dfrac{\pi }{7}}=-1\]