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Plzz ans both...................................more than one ootions are correct


one year ago

							secA = x + 1/4x As 1 + tan^2A = sec^2A tan^2A = sec^2A – 1   Therfore, tan^2A = (x + 1/4x)^2 – 1                                                                        = x^2 + 2*x*1/4x + 1/16x^2 – 1                                                                       = x^2 + 1/2 + 1/16x2 – 1                                                                        = x^2 + 1/16x^2 – 1/2                                                                        = (x – 1/4x)^2 Therefore, tan^2A = x – 1/4x or tan^2A = - (x – 1/4x) Substitute the value of secA and tanA in the given equation secA + tanA LHS = secA + tanA                          = x + 1/4x + x – 1/4x                           = 2x                           = RHS Or LHS = secA + tanA                         =x + 1/4x -x + 1/4x                            = 2/4x                            = 1/2x                          = RHS Hence proved.

one year ago
							A.    ;  ;   ;   ANS: Option 3​B. Given Substituting  value in above equation we get By slove this equation we getANS: Option 1

one year ago
							 secA = x + 1/4x As 1 + tan^2A = sec^2A tan^2A = sec^2A – 1   Therfore, tan^2A = (x + 1/4x)^2 – 1                                                                        = x^2 + 2*x*1/4x + 1/16x^2 – 1                                                                       = x^2 + 1/2 + 1/16x2 – 1                                                                        = x^2 + 1/16x^2 – 1/2                                                                        = (x – 1/4x)^2 Therefore, tan^2A = x – 1/4x or tan^2A = - (x – 1/4x) Substitute the value of secA and tanA in the given equation secA + tanA LHS = secA + tanA                          = x + 1/4x + x – 1/4x                           = 2x                           = RHS

9 months ago
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