Prove 2tan⁻¹{√(a−b)/√(a+b) tan(θ/2)} = cos⁻¹{(a cosθ + b)/(a + b cosθ)}

Prove that \( 2\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan\frac{\theta}{2}\right) = \cos^{-1}\left(\frac{a\cos\theta + b}{a + b\cos\theta}\right) \)

Solution:

Let

\[ t = \sqrt{\frac{a-b}{a+b}} \tan\frac{\theta}{2} \]

Then LHS becomes:

\[ 2\tan^{-1}(t) \]

Use identity:

\[ \cos(2\tan^{-1}t) = \frac{1 – t^2}{1 + t^2} \]

So,

\[ \cos(\text{LHS}) = \frac{1 – t^2}{1 + t^2} \]

Substitute \(t^2\):

\[ t^2 = \frac{a-b}{a+b} \tan^2\frac{\theta}{2} \]

Using identity:

\[ \tan^2\frac{\theta}{2} = \frac{1 – \cos\theta}{1 + \cos\theta} \]

\[ t^2 = \frac{a-b}{a+b} \cdot \frac{1 – \cos\theta}{1 + \cos\theta} \]

Now compute:

\[ \frac{1 – t^2}{1 + t^2} = \frac{(a+b)(1+\cos\theta) – (a-b)(1-\cos\theta)} {(a+b)(1+\cos\theta) + (a-b)(1-\cos\theta)} \]

Simplify numerator:

\[ = (a+b + a\cos\theta + b\cos\theta) – (a-b – a\cos\theta + b\cos\theta) \]

\[ = 2a\cos\theta + 2b \]

Denominator:

\[ = (a+b + a\cos\theta + b\cos\theta) + (a-b – a\cos\theta + b\cos\theta) \]

\[ = 2a + 2b\cos\theta \]

Thus,

\[ \cos(\text{LHS}) = \frac{a\cos\theta + b}{a + b\cos\theta} \]

Hence,

\[ \text{LHS} = \cos^{-1}\left(\frac{a\cos\theta + b}{a + b\cos\theta}\right) \]

which equals RHS.

\[ 2\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan\frac{\theta}{2}\right) = \cos^{-1}\left(\frac{a\cos\theta + b}{a + b\cos\theta}\right) \]

Spread the love