Trigonometric Identity Proof
Show that:
[NCERT, Ex. 2.3, Q.6, Page.35]
\[
2\tan^{-1}(-3) = -\frac{\pi}{2} + \tan^{-1}\left( -\frac{4}{3} \right)
\]
Solution
1
\[
\text{LHS} = 2\tan^{-1}(-3) = -2\tan^{-1}(3)
\]
Using the property: \(\tan^{-1}(-x) = -\tan^{-1}(x)\)
2
\[
= -\left[ \cos^{-1} \left( \frac{1 – 3^2}{1 + 3^2} \right) \right]
\]
Using the identity: \(2\tan^{-1}x = \cos^{-1} \left( \frac{1 – x^2}{1 + x^2} \right)\)
3
\[
= -\left[ \cos^{-1} \left( \frac{1 – 9}{1 + 9} \right) \right] = -\left[ \cos^{-1} \left( \frac{-8}{10} \right) \right] = -\left[ \cos^{-1} \left( -\frac{4}{5} \right) \right]
\]
Simplify the expression
4
\[
= -\left[ \pi – \cos^{-1} \left( \frac{4}{5} \right) \right]
\]
Using the property: \(\cos^{-1}(-x) = \pi – \cos^{-1}(x)\)
5
\[
= -\pi + \cos^{-1} \left( \frac{4}{5} \right)
\]
Distribute the negative sign
Let:
\[
\cos^{-1} \left( \frac{4}{5} \right) = \theta \Rightarrow \cos \theta = \frac{4}{5} \Rightarrow \tan \theta = \frac{3}{4} \Rightarrow \theta = \tan^{-1} \left( \frac{3}{4} \right)
\]
6
\[
= -\pi + \tan^{-1} \left( \frac{3}{4} \right)
\]
Substitute based on the definition above
7
\[
= -\pi + \left[ \frac{\pi}{2} – \cot^{-1} \left( \frac{3}{4} \right) \right]
\]
Using the identity: \(\tan^{-1}x = \frac{\pi}{2} – \cot^{-1}x\)
8
\[
= -\frac{\pi}{2} – \cot^{-1} \left( \frac{3}{4} \right)
\]
Simplify the expression
9
\[
= -\frac{\pi}{2} – \tan^{-1} \left( \frac{4}{3} \right)
\]
Using the identity: \(\cot^{-1}x = \tan^{-1} \left( \frac{1}{x} \right)\)
10
\[
= -\frac{\pi}{2} + \tan^{-1} \left( -\frac{4}{3} \right)
\]
Using the property: \(-\tan^{-1}x = \tan^{-1}(-x)\)
11
\[
= \text{RHS}
\]
\[
\therefore 2\tan^{-1}(-3) = -\frac{\pi}{2} + \tan^{-1}\left( -\frac{4}{3} \right)
\]