SSC SSC CHSL TIER-I Solved Paper Held on 11.01.2017

  • question_answer \[\tan \left( \frac{A}{2} \right)\]is equal to

    A)  \[\frac{\tan A}{(1+secA)}\]

    B)  \[\frac{1}{(cosecA+cotA)}\]

    C) \[\frac{\tan A}{(1+cosecA)}\]

    D) \[\frac{1}{(secA+\cot A)}\]

    Correct Answer: A

    Solution :

     \[\tan \frac{A}{2}=\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}\] \[=\frac{2\sin \frac{A}{2}.\cos \frac{A}{2}}{2{{\cos }^{2}}\frac{A}{2}}\] [Multiplying numerator and denominator by \[2\cos \frac{A}{2}\]] \[=\frac{\sin A}{1+\cos A}\] \[=\frac{\frac{\sin A}{\cos A}}{\frac{1+\cos A}{\cos A}}=\frac{\tan A}{\sec A+1}\] or When \[A={{60}^{o}},\] \[\tan \frac{A}{2}=\tan {{30}^{o}}=\frac{1}{\sqrt{3}}\] and \[\frac{\tan A}{1+\sec A}=\frac{\tan {{60}^{o}}}{1+\sec {{60}^{o}}}\] \[=\frac{\sqrt{3}}{1+2}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\]

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