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Question

Question: If we have the expression \[\tan \theta = a \ne 0\], \[\tan 2\theta = b \ne 0\] and \[\tan \theta + ...

If we have the expression tanθ=a0\tan \theta = a \ne 0, tan2θ=b0\tan 2\theta = b \ne 0 and tanθ+tan2θ=tan3θ\tan \theta + \tan 2\theta = \tan 3\theta then
(1)\left( 1 \right) a=ba = b
(2)\left( 2 \right) ab=1ab = 1
(3)\left( 3 \right) (a+b)=0\left( {a + b} \right) = 0
(4)\left( 4 \right) b=2ab = 2a

Explanation

Solution

We have to find the value of the given trigonometric expression tanθ+tan2θ=tan3θ\tan \theta + \tan 2\theta = \tan 3\theta . We solve this question using the concept of the various properties of the trigonometric functions. We should have the knowledge of the formula of the tangent of the sum of two angles. First we will simplify the right hand side of the given expression using the formula of the tangent of the sum of two angles and then we will simplify the expression and hence putting the values of the given tangent functions in the expression we will evaluate the required condition of the expression.

Complete step-by-step solution:
Given :
tanθ=a0\tan \theta = a \ne 0, tan2θ=b0\tan 2\theta = b \ne 0, tanθ+tan2θ=tan3θ\tan \theta + \tan 2\theta = \tan 3\theta
We know that the formula of the tangent of the sum of two angles is given as :
tan(a+b)=tana+tanb1tana×tanb\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a \times \tan b}}
We can write the right hand side as :
tan3θ=tan(θ+2θ)\tan 3\theta = \tan \left( {\theta + 2\theta } \right)
Now using the above formula, the right hand side of the expression can be written as :
tan3θ=tanθ+tan2θ1tanθ×tan2θ\tan 3\theta = \dfrac{{\tan \theta + \tan 2\theta }}{{1 - \tan \theta \times \tan 2\theta }}
Substituting the values in the right hand side, we get the expression as :
tanθ+tan2θ=tanθ+tan2θ1tanθ×tan2θ\tan \theta + \tan 2\theta = \dfrac{{\tan \theta + \tan 2\theta }}{{1 - \tan \theta \times \tan 2\theta }}
Now on simplify the terms, we can write the expression as :
(tanθ+tan2θ)(1tanθ×tan2θ)(tanθ+tan2θ)=0\left( {\tan \theta + \tan 2\theta } \right)\left( {1 - \tan \theta \times \tan 2\theta } \right) - \left( {\tan \theta + \tan 2\theta } \right) = 0
Taking (tanθ+tan2θ)\left( {\tan \theta + \tan 2\theta } \right) common, we get
(tanθ+tan2θ)(1tanθ×tan2θ1)=0\left( {\tan \theta + \tan 2\theta } \right)\left( {1 - \tan \theta \times \tan 2\theta - 1} \right) = 0
So, from the expression we can write the expression as :
(tanθ+tan2θ)=0\left( {\tan \theta + \tan 2\theta } \right) = 0 or (1tanθ×tan2θ1)=0\left( {1 - \tan \theta \times \tan 2\theta - 1} \right) = 0
(tanθ+tan2θ)=0\left( {\tan \theta + \tan 2\theta } \right) = 0 or (tanθ×tan2θ)=0\left( {\tan \theta \times \tan 2\theta } \right) = 0
Now, we will take the expression as :
(tanθ+tan2θ)=0\left( {\tan \theta + \tan 2\theta } \right) = 0
Putting the value in the expression, we get the value as :
(a+b)=0\left( {a + b} \right) = 0
Thus, we conclude that the expression for the given trigonometric expression tanθ+tan2θ=tan3θ\tan \theta + \tan 2\theta = \tan 3\theta is (a+b)=0\left( {a + b} \right) = 0.
Hence, the correct option is (3)\left( 3 \right).

Note: We will neglect the solution (tanθ×tan2θ)=0\left( {\tan \theta \times \tan 2\theta } \right) = 0 as it is only possible when either tanθ=0\tan \theta = 0 or tan2θ=0\tan 2\theta = 0 but it is given in the condition that tanθ0\tan \theta \ne 0 and tan2θ0\tan 2\theta \ne 0. So, as it contradicts the given conditions for the value we will neglect the solution.