3 Marks Question

Question 13 Marks

Find the value of $\tan \frac{1}{2}\left[ {{{\sin }^{ - 1}}\frac{{2x}}{{1 + {x^2}}} + {{\cos }^{ - 1}}\frac{{1 - {y^2}}}{{1 + {y^2}}}} \right]$, |x| < 1, y > 0 and xy < 1

Answer

Putting $x = \tan \theta$ and $y = \tan \phi$
$\therefore \tan \frac{1}{2}\left[ {{{\sin }^{ - 1}}\frac{{2x}}{{1 + {x^2}}} + {{\cos }^{ - 1}}\frac{{1 - {y^2}}}{{1 + {y^2}}}} \right]$
$ = \tan \frac{1}{2}\left[ {{{\sin }^{ - 1}}\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} + {{\cos }^{ - 1}}\frac{{1 - {{\tan }^2}\phi }}{{1 + {{\tan }^2}\phi }}} \right]$
$ = \tan \frac{1}{2}\left[ {{{\sin }^{ - 1}}\sin 2\theta + {{\cos }^{ - 1}}\cos 2\phi } \right]$
$ = \tan \frac{1}{2}\left[ {2\theta + 2\phi } \right]$
$ = \tan \left[ {\theta + \phi } \right]$
$= \frac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }}$
$ = \frac{{x + y}}{{1 - xy}}$

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Question 23 Marks

Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\frac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right),\;a > 0,\left( { - \frac{a}{{\sqrt 3 }} < x < \frac{a}{{\sqrt 3 }}} \right)$

Answer

${\tan ^{ - 1}}\left( {\frac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{3\left( {\frac{x}{a}} \right) - {{\left( {\frac{x}{a}} \right)}^3}}}{{1 - 3{{\left( {\frac{x}{a}} \right)}^2}}}} \right)$ $[$Dividing numerator and denominator by $a^3]$
Putting $\frac{x}{a} = \tan \theta$ so that $\theta = {\tan ^{ - 1}}\frac{x}{a}$
$ = {\tan ^{ - 1}}\left( {\frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right)$
$ = {\tan ^{ - 1}}\tan 3\theta$
$ = 3\theta = 3{\tan ^{ - 1}}\frac{x}{a}$

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MATHS 3 Marks Question

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