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STD 11 - 3. trignometrical ratios,functions and identities — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 3. trignometrical ratios,functions and identities4 Marks
MCQ
$\frac{{1 + \sin A - \cos A}}{{1 + \sin A + \cos A}} =$
  • A
    $\sin \frac{A}{2}$
  • B
    $\cos \frac{A}{2}$
  • ✓
    $\tan \frac{A}{2}$
  • D
    $\cot \frac{A}{2}$
✓

Answer

Correct option: C.
$\tan \frac{A}{2}$
c
(c) $\frac{{1 + \sin A - \cos A}}{{1 + \sin A + \cos A}}$

$ = \frac{{2\,{{\sin }^2}\frac{A}{2} + 2\,\sin \frac{A}{2}\cos \frac{A}{2}}}{{2\,{{\cos }^2}\frac{A}{2} + 2\,\sin \frac{A}{2}\cos \frac{A}{2}}}$

$ = \frac{{2\,\,\sin \frac{A}{2}\,\left( {\sin \frac{A}{2} + \cos \frac{A}{2}} \right)}}{{2\,\,\cos \frac{A}{2}\,\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} \right)}}$

$= \tan \frac{A}{2}$. 

Trick : Put $A = {60^o}.$ 

$\frac{{1 + (\sqrt 3 /2) - (1/2)}}{{1 + (\sqrt 3 /2) + (1/2)}} = \frac{{1 + \sqrt 3 }}{{3 + \sqrt 3 }} = \frac{1}{{\sqrt 3 }}$ 

which is given by option $(c)$,

$i.e.$ $\tan \frac{{{{60}^o}}}{2} = \frac{1}{{\sqrt 3 }}$

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