If y - tan -1-1 - cos x-1 - cos x and - - 227 - then the value of 7x - 14y3 - 73 is equal to

1+cos x1 cos x =2cos ^2x22sin ^2x2 =cos ^2x2sin ^2(x2) = ^2(x2) Hence tan ^ 1(√( ^2x2)) =tan ^ 1( (x2)) =π2 x2 Hence 2y=π ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

If y = tan ...

Question

If $y = {tan}^{- 1} \sqrt{\frac{1 + cos x}{1 - cos x}}$ and $π = \frac{22}{7}$ , then the value of $(7 x + 14 y)^{3} - 7^{3}$ is equal to

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Solution

$\frac{1 + cos x}{1 - cos x}$
$= \frac{2 {cos}^{2} \frac{x}{2}}{2 {sin}^{2} \frac{x}{2}}$
$= \frac{{cos}^{2} \frac{x}{2}}{{sin}^{2} (\frac{x}{2})}$
$= {cot}^{2} (\frac{x}{2})$
Hence
${tan}^{- 1} (\sqrt{{cot}^{2} \frac{x}{2}})$
$= {tan}^{- 1} (cot (\frac{x}{2}))$
$= \frac{π}{2} - \frac{x}{2}$
Hence
$2 y = π - x$
Hence
$2 y + x = π$
$14 y + 7 x = 7 π = 22$
Now
$(22)^{3} - 7^{3}$
$= 10648 - 343$
$= 10305$

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If y=tan−1√1+cosx1−cosx and π=227 , then the value of (7x+14y)3−73 is equal to

1+cosx1−cosx=2cos2x22sin2x2=cos2x2sin2(x2)=cot2(x2)Hencetan−1(√cot2x2)=tan−1(cot(x2))=π2−x2Hence2y=π−xHence2y+x=π14y+7x=7π=22Now (22)3−73=10648−343=10305

If $y = {tan}^{- 1} \sqrt{\frac{1 + cos x}{1 - cos x}}$ and $π = \frac{22}{7}$ , then the value of $(7 x + 14 y)^{3} - 7^{3}$ is equal to