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SSC Sample Paper SSC-CGL TIER - I Sample Test Paper-5

  • question_answer
    What is  \[\frac{{{\cos }^{2}}({{45}^{\circ }}+\theta )+co{{s}^{2}}({{45}^{\circ }}-\theta )}{\tan ({{60}^{\circ }}+\theta )tan({{30}^{\circ }}-\theta )}\] equal to?

    A) -1                                

    B) 0

    C) 1                                 

    D) 2

    Correct Answer: C

    Solution :

    \[\frac{{{\cos }^{2}}({{45}^{{}^\circ }}+\theta )+co{{s}^{2}}({{45}^{{}^\circ }}-\theta )}{\tan ({{60}^{{}^\circ }}+\theta )\tan ({{30}^{{}^\circ }}-\theta )}\] = \[\frac{\frac{\cos ({{90}^{{}^\circ }}+2\theta )+1}{2}+\frac{\cos ({{90}^{{}^\circ }}-2\theta )+1}{2}}{\tan ({{60}^{{}^\circ }}+\theta )\tan [{{90}^{{}^\circ }}-({{60}^{{}^\circ }}+\theta )}\] \[(\therefore \cos 2\theta =2{{\cos }^{2}}\theta -1)\] \[\frac{\frac{\cos ({{90}^{{}^\circ }}+2\theta )+\cos ({{90}^{{}^\circ }}-2\theta )}{2}+1}{\tan ({{60}^{{}^\circ }}+\theta )cot({{60}^{{}^\circ }}+\theta )}\] \[\frac{\frac{(-\sin 2\theta )+\sin 2\theta }{2}+1=\mathbf{1}}{1}\]


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