Question

Prove that $cos (2 tan^{-1} \frac{1}{7}) = sin (2 tan^{-1} \frac{3}{4})$.

Answer 1

Answer:

Step-by-step explanation:

$cos(2\:tan^{-1}\frac{1}{7} )=sin(2\:tan^{-1}\frac{3}{4})\\\\LHS\\\\2\:tan^{-1}x=tan^{-1}(\frac{2x}{1-x^{2} })\\\\(2\:tan^{-1}\frac{1}{7} )=tan^{-1}(\frac{\frac{2}{7} }{1-(\frac{1}{7}) ^{2} })\\\\=tan^{-1}(\frac{2\times49}{7\times48} )\\\\=tan^{-1}(\frac{7}{24})$

As we know that

$tan^{-1}x=cos^{-1}(\frac{1}{\sqrt{x^{2} +1} } )\\\\\\tan^{-1}(\frac{7}{24})=cos^{-1}(\frac{1}{\sqrt{(\frac{7}{24})^{2} +1} } )\\\\=cos^{-1}(\frac{24}{25})\\\\=> cos\:cos^{-1}(\frac{24}{25})\\\\=(\frac{24}{25})$...eq1

∴ cos and cos inverse cancels each other

RHS:

$tan^{-1}(\frac{2x}{1-x^{2} })\\\\(2\:tan^{-1}\frac{3}{4} )=tan^{-1}(\frac{\frac{3}{2} }{1-(\frac{3}{4}) ^{2} })\\\\\\=tan^{-1}(\frac{3\times16}{2\times7} )\\\\\\=tan^{-1}(\frac{24}{7})$

Also we know that

$tan^{-1}x=sin^{-1}(\frac{x}{\sqrt{x^{2} +1} } )\\\\\\tan^{-1}(\frac{24}{7})=sin^{-1}(\frac{\frac{24}{7}}{\sqrt{(\frac{24}{7})^{2} +1} } )\\\\=sin^{-1}(\frac{24}{25})\\\\=> sin\:sin^{-1}(\frac{24}{25})\\\\=(\frac{24}{25})$...eq2

from eq1 and eq2

LHS=RHS

hence proved

Answer 2

Answer:

Step-by-step explanation:

Formula used:

${tan}^{-1}A-{tan}^{-1}B={tan}^{-1}(\frac{A-B}{1+AB})$

$sin\theta=cos(\frac{\pi}{2}-\theta)$

$cos(2{tan^{-1}(\frac{1}{7}))$

$=sin(\frac{\pi}{2}-2{tan^{-1}(\frac{1}{7}))$

$=sin2(\frac{\pi}{4}-{tan^{-1}(\frac{1}{7}))$

$=sin2({tan^{-1}1-{tan^{-1}\frac{1}{7})$

$=sin2[{tan}^{-1}(\frac{1-\frac{1}{7}}{1+1.\frac{1}{7}})]$

$=sin2[{tan}^{-1}(\frac{\frac{6}{7}}{\frac{8}{7}})]$

$=sin2[{tan}^{-1}(\frac{6}{8})]$

$=sin2[{tan}^{-1}(\frac{3}{4})]$