Question

If m sin θ = n sin (θ + 2α) then tan (θ + α) cot α is equal to:
1. \(\frac{1\ -\ n}{1\ +\ n}\)
2. \(\frac{m\ +\ n}{m\ -\ n}\)
3. \(\frac{m\ -\ n}{m\ +\ n}\)
4. None of these

Answer 1

Correct Answer - Option 2 : \(\frac{m\ +\ n}{m\ -\ n}\)

Concept:

Componendo and dividendo rule:

If \(\frac{a}{b} = \frac{c}{d}\)

Then \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\)

Formula used:

1. sin C + sin D = 2sin(\(\frac{C\ +\ D}{2}\)) cos(\(\frac{C\ -\ D}{2}\))

2. sin C - sin D = 2sin(\(\frac{C\ -\ D}{2}\)) cos(\(\frac{C\ +\ D}{2}\))

3. \(\frac{sin\ \theta}{cos\ \theta}\ =\ tan\ \theta\)

4. \(\frac{cos\ \theta}{sin\ \theta}\ =\ cot\ \theta\)

Calculation:

Given that,

m sin θ = n sin (θ + 2α) then,

⇒ \(\frac{m}{n}\ =\ \frac{sin(\theta\ +\ 2\alpha)}{sin\ \theta}\)

Using componendo and dividendo rule

\(\frac{m\ +\ n}{m\ -\ n}\ =\ \frac{sin(\theta\ +\ 2\alpha)\ +\ sin\ \theta}{sin(\theta\ +\ 2\alpha)\ -\ sin\ \theta}\)

Using the identity discussed in the concept part,

⇒ \(\frac{m\ +\ n}{m\ -\ n}\ =\ \frac{2\ sin(\theta\ +\ \alpha)\ cos\ \alpha}{2\ cos(\theta\ +\ \alpha)\ sin\ \alpha}\)

⇒ \(\frac{m\ +\ n}{m\ -\ n}\ =\ tan(\theta\ +\ \alpha)cot\ \alpha\)