Trigonometric Proof
Prove that:
[NCERT, Ex. 2.3, Q.3, Page.35]
\[
\cot \left( \frac{\pi}{4} – 2\cot^{-1} 3 \right) = 7
\]
Solution
\[
\cot \left( \frac{\pi}{4} – 2\cot^{-1} 3 \right) = 7
\]
We are to prove this identity.
\[
\frac{\pi}{4} – 2\cot^{-1} 3 = \cot^{-1} 7
\]
Taking cot inverse on both sides (as cot is one-to-one in the principle range).
\[
2\cot^{-1} 3 = \frac{\pi}{4} – \cot^{-1} 7
\]
\[
2\tan^{-1} \frac{1}{3} = \frac{\pi}{4} – \tan^{-1} \frac{1}{7}
\]
Using the identity \(\cot^{-1} x = \tan^{-1} \frac{1}{x}\) for \(x > 0\).
\[
2\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{7} = \frac{\pi}{4}
\]
\[
\tan^{-1} \left( \frac{2 \cdot \frac{1}{3}}{1 – \left( \frac{1}{3} \right)^2} \right) + \tan^{-1} \frac{1}{7} = \frac{\pi}{4}
\]
Using the identity \(2\tan^{-1} x = \tan^{-1} \left( \frac{2x}{1 – x^2} \right)\) for \(|x| < 1\).
\[
\tan^{-1} \left( \frac{2/3}{8/9} \right) + \tan^{-1} \frac{1}{7} = \frac{\pi}{4}
\]
\[
\tan^{-1} \left( \frac{3}{4} \right) + \tan^{-1} \frac{1}{7} = \frac{\pi}{4}
\]
\[
\tan^{-1} \left( \frac{\frac{3}{4} + \frac{1}{7}}{1 – \left( \frac{3}{4} \cdot \frac{1}{7} \right)} \right) = \frac{\pi}{4}
\]
Using the identity \(\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A + B}{1 – AB} \right)\) (since \(AB < 1\)).
\[
\tan^{-1} \left( \frac{\frac{21 + 4}{28}}{\frac{28 – 3}{28}} \right) = \frac{\pi}{4}
\]
\[
\tan^{-1} \left( \frac{25}{25} \right) = \frac{\pi}{4}
\]
\[
\tan^{-1} (1) = \frac{\pi}{4}
\]
\[
\frac{\pi}{4} = \frac{\pi}{4}
\]
Hence, LHS = RHS. Proved.