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CBSE Class 10 Standard Math 2026 All Sets Solved Paper

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Question : 4 of 20
Marks: +1, -0
  1+tan2A1+cot2A\; \frac{1\, +\, \tan^2\, A}{1\, +\, \cot^2\, A} equals to:
Solution:  
  1+tan2A1+cot2A\; \frac{1 + \tan^2 A}{1 + \cot^2 A}
=  1+  sin2Acos2A1+  cos2sinA= \; \frac{1 + \; \frac{\sin^2 A}{\cos^2 A}}{1 + \; \frac{\cos^2}{\sin^A}}
=  cos2A+  sin2Acos2Asin2+  cos2Asin2A= \; \frac{\cos^2 A + \; \frac{\sin^2 A}{\cos^2 A}}{\sin^2 + \; \frac{\cos^2 A}{\sin^2 A}}
=    1cos2A  1sin2A= \; \frac{\; \frac{1}{\cos^2 A}}{\; \frac{1}{\sin^2 A}}
=  sin2Acos2A= \; \frac{\sin^2 A}{\cos^2 A}
=tan2A= \tan^2 A
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