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ICSE Class X Math 2015 Paper

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Question : 31 of 52
Marks: +1, -0
Prove that:
  sin⁡θ1−cot⁡θ+  cos⁡θ1−tan⁡θ\; \frac{\sin θ}{1-\cot θ} + \; \frac{\cos θ}{1-\tan θ} =cos⁡θ+sin⁡θ= \cos θ + \sin θ
Solution:  
Given :   sin⁡θ1−cot⁡θ+  cos⁡θ1−tan⁡θ\; \frac{\sin θ}{1-\cot θ} + \; \frac{\cos θ}{1-\tan θ} =cos⁡θ+sin⁡θ= \cos θ + \sin θ
L.H.S. =  sin⁡θ(1−cos⁡θsin⁡θ)+  cos⁡(1−sin⁡θcos⁡θ)= \; \frac{\sin θ}{\left(1 - \frac{\cos θ}{\sin θ}\right)} + \; \frac{\cos}{\left(1 - \frac{\sin θ}{\cos θ}\right)}
  =  sin⁡2θsin⁡θ−cos⁡θ+  cos⁡2θcos⁡θ−sin⁡θ\; = \; \frac{\sin^2 θ}{\sin θ - \cos θ} + \; \frac{\cos^2 θ}{\cos θ - \sin θ}
  =  sin⁡2θsin⁡θ−cos⁡θ−  cos⁡2θsin⁡θ−cos⁡θ\; = \; \frac{\sin^2 θ}{\sin θ - \cos θ} - \; \frac{\cos^2 θ}{\sin θ - \cos θ}
  =  sin⁡2θ−cos⁡2θsin⁡θ−cos⁡θ\; = \; \frac{\sin^2 θ - \cos^2 θ}{\sin θ - \cos θ}
  =  (sin⁡θ+cos⁡θ)(sin⁡θ−cos⁡θ)sin⁡θ−cos⁡θ\; = \; \frac{(\sin θ + \cos θ)(\sin θ - \cos θ)}{\sin θ - \cos θ}
  =sin⁡θ+cos⁡θ=  R.H.S.  \; = \sin θ + \cos θ = \; \text{R.H.S.} \;
L.H.S. = R.H.S.
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