Show that cos(2 tan⁻¹(1/7)) = sin(4 tan⁻¹(1/3)) | NCERT EXEMPLAR Class 12 Maths Ex 2.3 Q10

Solution

We have:

\[ \cos\left(2 \tan^{-1}\frac{1}{7}\right) = \sin\left(4 \tan^{-1}\frac{1}{3}\right) \]

Using formulas for double angle of arctan and arccos:

\[ \cos\Big[\cos^{-1}\frac{1-(1/7)^2}{1+(1/7)^2}\Big] = \sin\Big[2 \cdot 2 \tan^{-1}\frac{1}{3}\Big] \]

Simplifying fractions:

\[ \cos\Big[\cos^{-1}\frac{48/49}{50/49}\Big] = \sin\Big[2 \cdot \tan^{-1}\frac{2/3}{1-(1/3)^2}\Big] \]

\[ \cos\Big[\cos^{-1}\frac{48}{50}\Big] = \sin\Big[2 \tan^{-1}\frac{18}{24}\Big] \]

\[ \cos\Big[\cos^{-1}\frac{24}{25}\Big] = \sin\Big[2 \tan^{-1}\frac{3}{4}\Big] \]

\[ \sin\Big[2 \tan^{-1}\frac{3}{4}\Big] = \sin\Big[\sin^{-1}\frac{2 \cdot (3/4)}{1 + (3/4)^2}\Big] \]

\[ \frac{24}{25} = \sin\Big[\sin^{-1}\frac{3/2}{25/16}\Big] = \frac{24}{25} \]

Therefore, \(\boxed{\text{LHS = RHS}}\). Hence proved.