$\theta$ and $\alpha$ are two angles of the two different triangles, and $n$ is a real number. The sine of angle $\theta$ is equal to the product of $n$ and sine of sum of angles $\theta$ and $2 \alpha$.

$\sin \theta = n \sin(\theta + 2\alpha)$

On the basis of this equation, the value of $\tan(\theta +\alpha) \cot \alpha$ is required to find trigonometrically.

Directly, the equation cannot be transformed as the required form but it is possible if angles $\theta$ and $\theta + 2\alpha$ are added and also subtracted.

01

$\implies$ $\dfrac{\sin \theta}{\sin(\theta + 2\alpha)} = n$

02

The ratio of sine functions can be expressed in the form the ratio of sum to subtraction from by using the componendo and dividendo rule. It helps us to add and subtract the angles in next few steps.

$\implies$ $\dfrac{\sin \theta + \sin(\theta + 2\alpha)}{\sin \theta -\sin(\theta + 2\alpha)} = \dfrac{n+1}{n-1}$

03

The two sine functions with different angles are added in numerator. The sum of them can be expressed in product form by using the sum to product form transformation identity of sine functions.

Similarly, the two sine functions with same different angles are subtracted in denominator. The subtraction of them can also be expressed in product form by using sum of product identity in subtraction form of sine functions.

$\implies$ $\dfrac{2\sin \Bigg[\dfrac{\theta +\theta +2\alpha}{2}\Bigg] \cos \Bigg[\dfrac{\theta -(\theta +2\alpha)}{2}\Bigg]}{2\cos \Bigg[\dfrac{\theta +\theta +2\alpha}{2}\Bigg] \sin \Bigg[\dfrac{\theta -(\theta +2\alpha)}{2}\Bigg]}$ $=$ $\dfrac{n+1}{n-1}$

$\implies$ $\require{cancel} \dfrac{\cancel{2}\sin \Bigg[\dfrac{2\theta +2\alpha}{2}\Bigg] \cos \Bigg[\dfrac{\theta -\theta -2\alpha}{2}\Bigg]}{\cancel{2}\cos \Bigg[\dfrac{2\theta +2\alpha}{2}\Bigg] \sin \Bigg[\dfrac{\theta -\theta -2\alpha}{2}\Bigg]}$ $=$ $\dfrac{n+1}{n-1}$

$\implies$ $\require{\cancel} \dfrac{\sin \Bigg[\dfrac{2(\theta +\alpha)}{2}\Bigg] \cos \Bigg[\dfrac{\cancel{\theta} -\cancel{\theta} -2\alpha}{2}\Bigg]}{\cos \Bigg[\dfrac{2(\theta +\alpha)}{2}\Bigg] \sin \Bigg[\dfrac{\cancel{\theta} -\cancel{\theta} -2\alpha}{2}\Bigg]}$ $=$ $\dfrac{n+1}{n-1}$

$\implies$ $\require{\cancel} \dfrac{\sin \Bigg[\dfrac{\cancel{2}(\theta +\alpha)}{\cancel{2}}\Bigg] \cos \Bigg[\dfrac{-2\alpha}{2}\Bigg]}{\cos \Bigg[\dfrac{\cancel{2}(\theta +\alpha)}{\cancel{2}}\Bigg] \sin \Bigg[\dfrac{-2\alpha}{2}\Bigg]}$ $=$ $\dfrac{n+1}{n-1}$

$\implies$ $\require{\cancel} \dfrac{\sin(\theta +\alpha) \cos \Bigg[\dfrac{-\cancel{2}\alpha}{\cancel{2}}\Bigg]}{\cos(\theta +\alpha) \sin \Bigg[\dfrac{-\cancel{2}\alpha}{\cancel{2}}\Bigg]}$ $=$ $\dfrac{n+1}{n-1}$

$\implies$ $\dfrac{\sin(\theta +\alpha) \cos(-\alpha)}{\cos(\theta +\alpha) \sin(-\alpha)}$ $=$ $\dfrac{n+1}{n-1}$

04

Cosine function is having a negative angle in numerator and the sine function is also having negative angle in denominator. They can be expressed normally according to the trigonometric functions of allied angles identities.

$\implies$ $\dfrac{\sin(\theta +\alpha) \cos \alpha}{\cos(\theta +\alpha) (-\sin \alpha)} = \dfrac{n+1}{n-1}$

$\implies$ $\dfrac{\sin(\theta +\alpha) \cos \alpha}{-\cos(\theta +\alpha) \sin \alpha} = \dfrac{n+1}{n-1}$

$\implies$ $\dfrac{\sin(\theta +\alpha) \cos \alpha}{\cos(\theta +\alpha) \sin \alpha} = \dfrac{n+1}{-(n-1)}$

Split the fractional function as two multiplying factors as per the angle of the functions.

$\implies$ $\dfrac{\sin(\theta +\alpha)}{\cos(\theta +\alpha)} \times \dfrac{\cos \alpha}{\sin \alpha} = \dfrac{1+n}{1-n}$

05

According to quotient identities of the trigonometric functions. The quotient of sine of angle by cosine of angle is tangent and the quotient of cosine of angle by sine of angle is cotangent.

$\therefore \,\,\,\,\,\,$ $\tan(\theta +\alpha) \times \cot \alpha = \dfrac{1+n}{1-n}$

It is the required solution for this trigonometry problem and it can also be written as follows.

$\implies$ $\dfrac{\tan(\theta +\alpha)}{\tan \alpha} = \dfrac{1+n}{1-n}$

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