11th Class Mathematics Trigonometric Identities Question Bank Critical Thinking

  • question_answer \[\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\] is equal to [IIT 1966, 1975]

    A) \[\cot 7\frac{{{1}^{o}}}{2}\]

    B) \[\sin 7\frac{{{1}^{o}}}{2}\]

    C) \[\sin \,{{15}^{o}}\]

    D) \[\cos \,\,{{15}^{o}}\]

    Correct Answer: A

    Solution :

    We have \[\cot A=\frac{\cos A}{\sin A}=\frac{2{{\cos }^{2}}A}{2\sin A\cos A}=\frac{1+\cos 2A}{\sin 2A}\] Putting \[A=7\frac{{{1}^{o}}}{2}\Rightarrow \cot 7\frac{{{1}^{o}}}{2}=\frac{1+\cos {{15}^{o}}}{\sin {{15}^{o}}}\] On simplification, we get \[\cot 7\frac{{{1}^{o}}}{2}=\sqrt{6}+\sqrt{2}+\sqrt{3}+\sqrt{4}\].

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