JEE Main & Advanced Mathematics Trigonometric Identities Question Bank Trigonometrical ratios of multiple and sub multiple angles

  • question_answer
    If \[\sin A+\cos A=\sqrt{2},\]then \[{{\cos }^{2}}A=\]

    A) \[\frac{1}{4}\]

    B) \[\frac{1}{2}\]

    C) \[\frac{1}{\sqrt{2}}\]

    D) \[\frac{3}{2}\]

    Correct Answer: B

    Solution :

    \[\sin A+\cos A=\sqrt{2}\]. On squaring both the sides Þ \[1+\sin 2A=2\,\Rightarrow \sin 2A=1=\sin {{90}^{o}}\] Þ \[2A={{90}^{o}}\]or \[A={{45}^{o}}\] Now, \[{{\cos }^{2}}A={{(\cos {{45}^{o}})}^{2}}={{\left( \frac{1}{\sqrt{2}} \right)}^{2}}=\frac{1}{2}\].

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