InterviewSolution
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1). 12). -13). 04). 2 |
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Answer» $(\frac{{\left[ {4cos\left( {90 - A} \RIGHT).si{n^3}\left( {90\; + \;A} \right)} \right] - \left[ {4sin\left( {90\; + \;A} \right).CO{s^3}\left( {90 - A} \right)} \right]}}{{cos\left( {\frac{{180\; + \;8A}}{2}} \right)\;}})$ $(\Rightarrow \frac{{\left( {4sinA.co{s^3}A} \right) - \left( {4cosA.si{n^3}A} \right)}}{{ - sin4A}})$ $(\Rightarrow \frac{{\left( {4sinA.co{s^3}A} \right) - \left( {4cosA.si{n^3}A} \right)}}{{ - 2sin2Acos2A}})$ $(\Rightarrow \frac{{\left( {4sinA.co{s^3}A} \right) - \left( {4cosA.si{n^3}A} \right)}}{{ - 4sinA.cosA.COS2A}})$ $(\Rightarrow \frac{{co{s^2}A - si{n^2}A\;}}{{ - cos2A}})$ Since, cos2A = cos2A – sin2A $(\Rightarrow \frac{{co{s^2}A - si{n^2}A\;}}{{-\left( {co{s^2}A-si{n^2}A} \right)}})$ ⇒ -1 |
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