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

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 3. trignometrical ratios,functions and identities1 Mark
MCQ
Suppose $\theta $ and $\phi  (\ne 0)$ are such that $sec\,(\theta  + \phi ),$ $sec\,\theta $ and $sec\,(\theta  - \phi )$ are in $A.P.$ If $cos\,\theta  = k\,cos\,( \frac {\phi }{2})$ for some $k,$ then $k$ is equal to
  • ✓
    $ \pm \sqrt 2 $
  • B
    $ \pm  1 $
  • C
    $ \pm \frac{1}{{\sqrt 2 }}$
  • D
    $ \pm  2 $
✓

Answer

Correct option: A.
$ \pm \sqrt 2 $
a
Since, $\sec \,(\theta  - \phi ),\,\,\sec \,\theta $ and $\sec \,(\theta  + \phi )$ are in $A.P.,$

$\therefore \,2\,\sec \,\theta \, = \,\sec \,(\theta  - \phi ) + \sec \,(\theta  + \phi )$

$ \Rightarrow \frac{2}{{\cos \theta \,}} = \frac{{\cos \,(\theta  + \phi ) + \,\cos \,(\theta  - \phi )}}{{\cos \,(\theta  - \phi )\,\cos \,(\theta  + \phi )}}$

$ \Rightarrow \,2({\cos ^2}\theta  - {\sin ^2} \phi  )\, = \,\cos \,\theta \,[2\,\cos \theta \,\cos \phi ]$

$ \Rightarrow \,{\cos ^2}\theta \,(1 - \cos \phi )\, = \,{\sin ^2}\phi \, = \,1 - {\cos ^2}\phi $

$ \Rightarrow \,{\cos ^2}\theta \, = 1 + \cos \phi \, = 2{\cos ^2}\frac{\phi }{2}$

$\therefore \,\,\cos \,\theta \, = \, \pm \,\sqrt 2 \cos \frac{\phi }{2}$

But given $\cos \,\theta \, = k\cos \frac{\phi }{2}$

$\therefore \,\,k = \,\, \pm \,\sqrt 2 $

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