Prove that $\cot \left( {\frac{\pi }{4} – 2{{\cot }^{ – 1}}3} \right) = 7$.
[NCERT,Exemplar.2.3,Q.3,Page.35]

Prove that $\cot \left( {\frac{\pi }{4} – 2{{\cot }^{ – 1}}3} \right) = 7$.
[NCERT,Exemplar.2.3,Q.3,Page.35]

We have to prove, $\cot \left( {\frac{\pi }{4} – 2{{\cot }^{ – 1}}3} \right) = 7$

$ \Rightarrow \left( {\frac{\pi }{4} – 2{{\cot }^{ – 1}}3} \right) = {\cot ^{ – 1}}7$

$ \Rightarrow \left( {2{{\cot }^{ – 1}}3} \right) = \frac{\pi }{4} – {\cot ^{ – 1}}7$

$ \Rightarrow $$2{\tan ^{ – 1}}\frac{1}{3} = \frac{\pi }{4} – {\tan ^{ – 1}}\frac{1}{7}$

$ \Rightarrow $$2{\tan ^{ – 1}}\frac{1}{3} + {\tan ^{ – 1}}\frac{1}{7} = \frac{\pi }{4}$


$ \Rightarrow $${\tan ^{ – 1}}\frac{{2/3}}{{1 – {{(1/3)}^2}}} + {\tan ^{ – 1}}\frac{1}{7} = \frac{\pi }{4}$

$ \Rightarrow $${\tan ^{ – 1}}\frac{{2/3}}{{8/9}} + {\tan ^{ – 1}}\frac{1}{7} = \frac{\pi }{4}$


$ \Rightarrow $${\tan ^{ – 1}}\frac{3}{4} + {\tan ^{ – 1}}\frac{1}{7} = \frac{\pi }{4}$

$ \Rightarrow $${\tan ^{ – 1}}\frac{{\frac{3}{4} + \frac{1}{7}}}{{1 – \frac{3}{4} \cdot \frac{1}{7}}} = \frac{\pi }{4}$

$ \Rightarrow $${\tan ^{ – 1}}\frac{{(21 + 4)/28}}{{(28 – 3)/28}} = \frac{\pi }{4}$
$ \Rightarrow $${\tan ^{ – 1}}\frac{{25}}{{25}} = \frac{\pi }{4}$

$ \Rightarrow $$1 = \tan \frac{\pi }{4}$

$ \Rightarrow $$1 = 1$
$ \Rightarrow $${\rm{LHS}} = {\rm{RHS}}$

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