SSC Quantitative Aptitude Trigonometry Question Bank Trigonometry (II)

  • question_answer
                 If \[\sin \,(10{}^\circ 6'32')=a,\] then the value of \[\cos \,(79{}^\circ 53'28'')+tan\,(10{}^\circ 6'32'')\]is

    A) \[\frac{a\,(1+\sqrt{1-{{a}^{2}}})}{\sqrt{1-{{a}^{2}}}}\]

    B) \[\frac{1+\sqrt{1-{{a}^{2}}}}{\sqrt{1-{{a}^{2}}}}\]

    C) \[\frac{\sqrt{1-{{a}^{2}}}+a}{\sqrt{1-{{a}^{2}}}}\]

    D) \[\frac{a\sqrt{1-{{a}^{2}}}+1}{\sqrt{1-{{a}^{2}}}}\]

    Correct Answer: A

    Solution :

    [a] \[\sin \,(10{}^\circ 6'32'')=a\] \[\sin \,(90{}^\circ -79{}^\circ 53'28'')=a\] \[\cos 79{}^\circ 53'28''=a\] \[\therefore \]      \[\cos \,(79{}^\circ 53'28'')+tan\,(10{}^\circ 6'32'')\]             \[a+\frac{a}{\sqrt{1-{{a}^{2}}}}=\frac{a(1+\sqrt{1-{{a}^{2}}})}{\sqrt{1-{{a}^{2}}}}\]


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