Question

Prove that cot (π/4 – 2 cot-1 3) = 7​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{cot\left(\dfrac{\pi}{4}-2\,cot^{-1}(3)\right)}$

$\tt{=\dfrac{cot\left(2\,cot^{-1}(3)\right)\cdot\,cot\left(\dfrac{\pi}{4}\right)+1}{cot\left(2\,cot^{-1}(3)\right)-cot\left(\dfrac{\pi}{4}\right)}}$

$\tt{=\dfrac{cot\left\{2\,tan^{-1}\left(\dfrac{1}{3}\right)\right\}\cdot\,1+1}{cot\left\{2\,tan^{-1}\left(\dfrac{1}{3}\right)\right\}-1}}$

$\tt{=\dfrac{cot\left\{tan^{-1}\left(\dfrac{\dfrac{2}{3}}{1-\dfrac{1}{9}}\right)\right\}+1}{cot\left\{tan^{-1}\left(\dfrac{\dfrac{2}{3}}{1-\dfrac{1}{9}}\right)\right\}-1}}$

$\tt{=\dfrac{cot\left\{tan^{-1}\left(\dfrac{2\times3}{9-1}\right)\right\}+1}{cot\left\{tan^{-1}\left(\dfrac{2\times3}{9-1}\right)\right\}-1}}$

$\tt{=\dfrac{cot\left\{tan^{-1}\left(\dfrac{6}{8}\right)\right\}+1}{cot\left\{tan^{-1}\left(\dfrac{6}{8}\right)\right\}-1}}$

$\tt{=\dfrac{cot\left\{tan^{-1}\left(\dfrac{3}{4}\right)\right\}+1}{cot\left\{tan^{-1}\left(\dfrac{3}{4}\right)\right\}-1}}$

$\tt{=\dfrac{cot\left\{cot^{-1}\left(\dfrac{4}{3}\right)\right\}+1}{cot\left\{cot^{-1}\left(\dfrac{4}{3}\right)\right\}-1}}$

$\tt{=\dfrac{\dfrac{4}{3}+1}{\dfrac{4}{3}-1}}$

$\tt{=\dfrac{4+3}{4-3}}$

$\tt{=\dfrac{7}{1}}$

$\tt{=7}$

Answer 2

Answer:

Hello dear.....

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