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Example – 5. If x = r cosαsinβ, y = rcosαcosβ and z = rsinα, show that x2 + y2 + z2 = r2.

Solution :-  x2 + y2 + z2 = r2cos2αsin2β + r2cos2αcos2β + r2sin2α
                                        = r2(cos2αsin2β + cos2αcos2β + sin2α)
                                        = r2[cos2α(sin2β + cos2β) + sin2α]
                                        = r2(cos2α + sin2α)                   [sin2β + cos2β= 1]
                                        = r2                                            [cos2α + sin2α = 1]

 

Example – 6. Prove that : (cotθ – cosecθ)2 = (1 – cosθ)/(1 + cosθ)

Solution :-  LHS = (cotθ – cosecθ)2
                            = (cosθ/sinθ – 1/sinθ)2
                            = [(cosθ - 1)/sinθ]2
                            = (1 – cosθ )2/sin2θ
                            =  (1 – cosθ)2/(1 – cos2θ)
                            = (1 – cosθ )2/(1 – cosθ )(1 + cosθ) 
                            = (1 – cosθ)/(1 + cosθ) 
                            = RHS

Example - 7. Without using trigonometric table, prove that :

(cos220º + cos270º)/(sin220º + sin270º) + sin264º + cos64º.sin26º = 2.

Solution :- LHS = (cos220º + cos270º)/(sin220º + sin270º) + sin264º + cos64º.sin26º
                          = [ cos220º + cos2(90º – 20º)]/[sin220º + sin2(90º – 20º)] + sin264º + cos64º.sin(90º – 64º) 
                            = (cos220º + sin220º)/(sin220º + cos220º) + sin264º + cos64º.cos64º
                                                                              [cos(90º – θ) = sinθ; sin(90º – θ) = cosθ]
                            = 1 + (sin264º + cos264º)          [sin2θ + cos2θ = 1]
                            = 1 + 1 = 2 = RHS

 

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Maths by Mr. M. P. Keshari

 

Chapter - 24   Chapter - 25   Chapter - 26
  Chapter - 27   Chapter - 28   Chapter - 29

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