InterviewSolution
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InterviewSolution
| 1. |
(Sina/1+cosa+1+cosa/sina)(cosa/1+sina+1+sina/cosa)=4seca×coseca |
| Answer» LHS=SinA/1+CosA+1+CosA/SinA(CosA/1+SinA+1+SinA/CosA)=Sin2+(1+CosA)2/SinA(1+CosA){Cos2A+(1+SinA)2/CosA(1+SinA)}=Sin2A(1+Cos2A+2CosA)/SinA+SinA×CosA(Cos2A+1+Sin2A+2SinA/CosA+SinA×CosA)=1+Sin2A+Cos2A+2CosA/SinA+SinA×CosA(1+Sin2A+Cos2A+2SinA/CosA+SinA×CosA)=2+2CosA/SinA(1+CosA){2+2SinA/CosA(1+SinA)} (Formula used : Sin2A+Cos2A = 1)=2(1+CosA)/SinA(1+CosA){2(1+SinA)/CosA(1+SinA}=2×1/SinA×1(2×1/CosA×1)=2/SinA(2/CosA)=2CosecA×2SecA (Formula used : 1- 1/SinA = CosecA 2- 1/CosA = SecA)=4CosecA×SecA So, L.H.S = R.H.S | |