Question

12. tan^2 A - sin^2 A = tan’^2A . sin^2 A​

Answer 1

Answer:

Let us first find the value of left hand side (LHS) that is tan2A−sin2A as shown below:

       

tan2A−sin2A=cos2Asin2A−sin2A(∵tanx=cosxsinx)=cos2Asin2A−sin2Acos2A=sin2A(cos2A1−cos2A)=cos2Asin2A(1−cos2A)=tan2Asin2A=RHS(∵tanx=cosxsinx,1−cos2x=sin2x)    

 

Since LHS=RHS,

      

Hence, 

Answer 2

Answer:

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