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ICSE Class X Math 2016 Paper

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Question : 51 of 57
Marks: +1, -0
Prove that :   cos⁡A1+sin⁡  A+tan⁡A=sec⁡A\;\frac{\cos A}{1+ \sin \;A} + \tan A = \sec A .
Solution:  
   L.H.S.     =  cos⁡A1+sin⁡  A+tan⁡A\;\text{ L.H.S. }\;\;=\;\frac{\cos A}{1+ \sin \; A} + \tan A
  =  cos⁡A1+sin⁡  A+  sin⁡  Acos⁡A\;=\;\frac{\cos A}{1+ \sin \; A} + \; \frac{\sin \; A}{\cos A}
  =  cos⁡2A+sin⁡  A(1+sin⁡  A)(1+sin⁡  A)cos⁡A\;=\;\frac{\cos^2 A + \sin \; A (1+ \sin \; A)}{(1+ \sin \; A) \cos A}
  =  cos⁡2A+sin⁡  A+sin⁡2A(1+sin⁡  A)(cos⁡A)\;=\;\frac{\cos^2 A + \sin \; A + \sin^2 A}{(1+ \sin \; A)(\cos A)}
  =  1+sin⁡  A(1+sin⁡  A)cos⁡A\;=\;\frac{1+ \sin \; A}{(1+ \sin \; A) \cos A}
  =  1cos⁡A\;=\;\frac{1}{\cos A}
  =sec⁡A=   R.H.S.   \;=\sec A = \;\text{ R.H.S. }\;
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