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Pls answer the above question with easy stepUnderstandable thanks in advance

Pls answer the above question with easy stepUnderstandable thanks in advance

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Grade:12th pass

2 Answers

Amit
405 Points
5 years ago
Hi there.. for solving divide by √a²+b² inside the log. Write a/√a²+b² as cosx hence b/√a²+b² becomes Sinx. Write the numerator as e^-ix and denominator as e^ix.. it becomes e^-2ix.. now use log property to get the angle inside cos as -2ix ×i which simplify to cos2x..use cos2x as 2cos²x -1.. Put value of cosx to get answer.. please approve..
Samyak Jain
333 Points
5 years ago
(a – ib)}]Let a = r cosA , b = r sin A,  where A is some angle & r is magnitude of either (a + ib) or (a – ib)\slashTo find : cos[i log{(a + ib)
(cos A + i sin A)\slashHere tan A = b / a(a -ib) / (a+ib) = (cos A - i sin A)   =  i( – 2Ai) *1 ( –2Ai)(a – ib)} = i log e\slash i log{(a + ib)\therefore   ( –2Ai) 1     = cos 2A - i sin 2A = e\slashA – 2i sin A cos A) 2  A – sin2= – 1] = ( cos2  i\becauseA) [2  A + sin2 (cos\slash2 = ( cos A - i sin A )
= 2A
cos [i log{(a + ib)\slash(a – ib)}]  =  cos 2A = (1-tan2 A) \slash (1 + tan2 A)
=  (1 – b2\slasha2) \slash (1+ b2\slasha2 )
cos [i log{(a + ib)\slash(a – ib)}]  =  (a2 – b2) \slash (a2 + b2)

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