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STD 12 - 2. inverse trigonometric function — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. inverse trigonometric function4 Marks
MCQ
$2{\tan ^{ - 1}}\left[ {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{\theta }{2}} \right] = $
  • ✓
    ${\cos ^{ - 1}}\left( {\frac{{a\cos \theta + b}}{{a + b\cos \theta }}} \right)$
  • B
    ${\cos ^{ - 1}}\left( {\frac{{a + b\cos \theta }}{{a\cos \theta + b}}} \right)$
  • C
    ${\cos ^{ - 1}}\left( {\frac{{a\cos \theta }}{{a + b\cos \theta }}} \right)$
  • D
    ${\cos ^{ - 1}}\left( {\frac{{a\cos + b\theta }}{{a + b\cos \theta }}} \right)$
✓

Answer

Correct option: A.
${\cos ^{ - 1}}\left( {\frac{{a\cos \theta + b}}{{a + b\cos \theta }}} \right)$
a
(a) $2{\tan ^{ - 1}}\left[ {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{\theta }{2}} \right]$

$ = {\cos ^{ - 1}}\left[ {\frac{{1 - \left( {\frac{{a - b}}{{a + b}}} \right){{\tan }^2}\frac{\theta }{2}}}{{1 + \left( {\frac{{a - b}}{{a + b}}} \right){{\tan }^2}\frac{\theta }{2}}}} \right]$

$\left( {\because 2{{\tan }^{ - 1}}x = {{\cos }^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$

$ = {\cos ^{ - 1}}\left[ {\frac{{(a + b) - (a - b){{\tan }^2}\frac{\theta }{2}}}{{(a + b) + (a - b){{\tan }^2}\frac{\theta }{2}}}} \right]$

$ = {\cos ^{ - 1}}\left[ {\frac{{a\left( {1 - {{\tan }^2}\frac{\theta }{2}} \right) + b\left( {1 + {{\tan }^2}\frac{\theta }{2}} \right)}}{{a\left( {1 + {{\tan }^2}\frac{\theta }{2}} \right) + b\left( {1 - {{\tan }^2}\frac{\theta }{2}} \right)}}} \right]$

$ = {\cos ^{ - 1}}\left[ {\frac{{\frac{{a\left( {1 - {{\tan }^2}\frac{\theta }{2}} \right)}}{{1 + {{\tan }^2}\frac{\theta }{2}}} + b}}{{a + b\left( {\frac{{1 - {{\tan }^2}\frac{\theta }{2}}}{{1 + {{\tan }^2}\frac{\theta }{2}}}} \right)}}} \right] $

$= {\cos ^{ - 1}}\left[ {\frac{{a\cos \theta + b}}{{a + b\cos \theta }}} \right]$.

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