
Solve the following equation for \[0<\theta <\dfrac{\pi }{2}\] :
\[\tan \theta +\tan 2\theta +\tan 3\theta =0\]
A. \[\tan \theta =0\]
B. \[\tan 2\theta =0\]
C. \[\tan 3\theta =0\]
D. \[\tan \theta \tan 2\theta =2\]
Answer
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Hint: Split the \[\tan 3\theta \] by using the formula \[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\] where A= \[2\theta \] and B= \[\theta \] here A and B values are taken like that because the problem is in terms of \[\theta \] , \[2\theta \] . After substituting the obtained expression in the given expression then we will get two values.
Complete step-by-step answer:
Given that \[0 < \theta < \dfrac{\pi }{2}\]
\[\tan \theta +\tan 2\theta +\tan 3\theta =0\] . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We can write the \[\tan 3\theta \] as follows
\[\tan 3\theta =\tan \left( 2\theta +\theta \right)\]
\[\Rightarrow \tan 3\theta =\dfrac{\tan 2\theta +\tan \theta }{1-\tan 2\theta \tan \theta }\]
Cross multiplying the above equation we will get as follows,
\[\Rightarrow \tan 3\theta -\tan 2\theta \tan \theta \tan 3\theta =\tan 2\theta +\tan \theta \] . . . . . . . . . . . . . . . . . . . . . (2)
Now substitute the value of equation (2) in the equation (1) then we will get,
\[\Rightarrow \tan \theta +\tan 2\theta +\tan 3\theta =0\]
\[\Rightarrow \tan 3\theta -\tan 2\theta \tan \theta \tan 3\theta +\tan 3\theta =0\]
\[\Rightarrow 2\tan 3\theta -\tan 2\theta \tan \theta \tan 3\theta =0\]
Now take \[\tan 3\theta \] common from each term then we will get,
\[\Rightarrow \tan 3\theta \left[ 2-\tan 2\theta \tan \theta \right] =0\] . . . . . . . . . . . . . . . . . (3)
Case-1
\[\tan 3\theta =0\] . . . . . . . . . . . . (4)
Case-2
\[2-\tan 2\theta \tan \theta =0\]
\[\Rightarrow \tan \theta \tan 2\theta =2\] . . . . . . . . . . . . . . (5)
So the correct option for above question is option (C) and option (D).
Note: To solve this type of problems we have to know trigonometric formulas like formula for expansion of \[\tan \left( A+B \right)\] . In trigonometric problems first we have to remember all formulas related to the problem and use the formula in which we will get the required answer. Generally by seeing the problem we will understand the approach to the problem.
Complete step-by-step answer:
Given that \[0 < \theta < \dfrac{\pi }{2}\]
\[\tan \theta +\tan 2\theta +\tan 3\theta =0\] . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We can write the \[\tan 3\theta \] as follows
\[\tan 3\theta =\tan \left( 2\theta +\theta \right)\]
\[\Rightarrow \tan 3\theta =\dfrac{\tan 2\theta +\tan \theta }{1-\tan 2\theta \tan \theta }\]
Cross multiplying the above equation we will get as follows,
\[\Rightarrow \tan 3\theta -\tan 2\theta \tan \theta \tan 3\theta =\tan 2\theta +\tan \theta \] . . . . . . . . . . . . . . . . . . . . . (2)
Now substitute the value of equation (2) in the equation (1) then we will get,
\[\Rightarrow \tan \theta +\tan 2\theta +\tan 3\theta =0\]
\[\Rightarrow \tan 3\theta -\tan 2\theta \tan \theta \tan 3\theta +\tan 3\theta =0\]
\[\Rightarrow 2\tan 3\theta -\tan 2\theta \tan \theta \tan 3\theta =0\]
Now take \[\tan 3\theta \] common from each term then we will get,
\[\Rightarrow \tan 3\theta \left[ 2-\tan 2\theta \tan \theta \right] =0\] . . . . . . . . . . . . . . . . . (3)
Case-1
\[\tan 3\theta =0\] . . . . . . . . . . . . (4)
Case-2
\[2-\tan 2\theta \tan \theta =0\]
\[\Rightarrow \tan \theta \tan 2\theta =2\] . . . . . . . . . . . . . . (5)
So the correct option for above question is option (C) and option (D).
Note: To solve this type of problems we have to know trigonometric formulas like formula for expansion of \[\tan \left( A+B \right)\] . In trigonometric problems first we have to remember all formulas related to the problem and use the formula in which we will get the required answer. Generally by seeing the problem we will understand the approach to the problem.
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