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Show that (2^{-1} 17 ) = (4^{-1} 13 ).

Question

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

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Show that

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

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Text solutionVerified

We have,

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

.

L H S

= cos ⎣ ⎡ cos^{- 1} ⎝ ⎛ \frac{1 - ( \frac{1}{7} ) ^{2}}{1 + ( \frac{1}{7} ) ^{2}} ⎠ ⎞ ⎦ ⎤

= cos ⎣ ⎡ cos^{- 1} ⎝ ⎛ \frac{\frac{48}{48}}{\frac{50}{49}} ⎠ ⎞ ⎦ ⎤

= cos [cos^{- 1} (\frac{48 \times 49}{50 \times 49})]

= cos [cos^{- 1} (\frac{24}{25})]

= cos [cos^{- 1} (\frac{24}{25})]

= \frac{24}{25}

R H S

sin (2.2 tan^{- 1} \frac{1}{3}) [∵ 2 tan^{- 1} x = cos^{_{1}} (\frac{1 - x ^{2}}{1 + x ^{2}})]

= sin ⎣ ⎡ 2. ⎝ ⎛ tan^{- 1} \frac{\frac{2}{3}}{1 - ( \frac{1}{3} ) ^{2}} ⎠ ⎞ ⎦ ⎤

= sin [2 tan^{- 1} (\frac{18}{28})]

= sin (2 tan^{- 1} \frac{2}{3})

= sin ⎝ ⎛ sin^{- 1} \frac{2 \times \frac{3}{4}}{1 + \frac{9}{16}} ⎠ ⎞ [∵ 2 tan^{- 1} x = sin^{- 1} \frac{2 x}{1 + x ^{2}}]

= sin (sin^{- 1} \frac{3/2}{25/16})

= \frac{48}{50} = \frac{24}{25}

∵ L H S = R H S

Hence proved

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Question Text	Show that $cos (2 tan^{- 1} \frac{1}{7}) = sin (4 tan^{- 1} \frac{1}{3})$ .
Answer Type	Text solution:1
Upvotes	150