11th Class Mathematics Trigonometric Identities Question Bank Critical Thinking

  • question_answer If \[\left| \cos \,\theta \,\left\{ \sin \theta +\sqrt{{{\sin }^{2}}\theta +{{\sin }^{2}}\alpha } \right\}\, \right|\,\le k,\] then the value of k is

    A) \[\sqrt{1+{{\cos }^{2}}\alpha }\]

    B) \[\sqrt{1+{{\sin }^{2}}\alpha }\]

    C) \[\sqrt{2+{{\sin }^{2}}\alpha }\]

    D) \[\sqrt{2+{{\cos }^{2}}\alpha }\]

    Correct Answer: B

    Solution :

    Let \[u=\cos \theta \left\{ \sin \theta +\sqrt{{{\sin }^{2}}\theta +{{\sin }^{2}}\alpha } \right\}\] Þ \[{{(u-\sin \theta \cos \theta )}^{2}}={{\cos }^{2}}\theta ({{\sin }^{2}}\theta +{{\sin }^{2}}\alpha )\] Þ \[{{u}^{2}}{{\tan }^{2}}\theta -2u\tan \theta +{{u}^{2}}-{{\sin }^{2}}\alpha =0\] Since tan \[\theta \]is real, therefore Þ \[4{{u}^{2}}-4{{u}^{2}}({{u}^{2}}-{{\sin }^{2}}\alpha )\ge 0\] \[\Rightarrow {{u}^{2}}-(1+{{\sin }^{2}}\alpha )\le 0\] Þ \[|u|\,\le \sqrt{1+{{\sin }^{2}}\alpha }\].

You need to login to perform this action.
You will be redirected in 3 sec spinner