Solution : The value of cosec 45 degrees is \(\sqrt{2}\). Proof : Let ABC be a triangle, right angled at B, in which \(\angle\) A = \(\angle\) C = 45 degrees \(\therefore\) BC = AB Let AB = BC = a Then by pythagoras theorem, \(AC^2\) = \(AB^2\) + \(BC^2\) = \(a^2\) + \(a^2\) = \(2a^2\) …
Solution : The value of sec 45 degrees is \(\sqrt{2}\). Proof : Let ABC be a triangle, right angled at B, in which \(\angle\) A = \(\angle\) C = 45 degrees \(\therefore\) BC = AB Let AB = BC = a Then by pythagoras theorem, \(AC^2\) = \(AB^2\) + \(BC^2\) = \(a^2\) + \(a^2\) = …
Solution : The value of cot 45 degrees is 1. Proof : Let ABC be a triangle, right angled at B, in which \(\angle\) A = \(\angle\) C = 45 degrees \(\therefore\) BC = AB Let AB = BC = a Then by pythagoras theorem, \(AC^2\) = \(AB^2\) + \(BC^2\) = \(a^2\) + \(a^2\) = …
Solution : The value of sin 45 degrees is \(1\over \sqrt{2}\). Proof : Let ABC be a triangle, right angled at B, in which \(\angle\) A = \(\angle\) C = 45 degrees \(\therefore\) BC = AB Let AB = BC = a Then by pythagoras theorem, \(AC^2\) = \(AB^2\) + \(BC^2\) = \(a^2\) + \(a^2\) …
Solution : The value of cos 45 degrees is \(1\over \sqrt{2}\). Proof : Let ABC be a triangle, right angled at B, in which \(\angle\) A = \(\angle\) C = 45 degrees \(\therefore\) BC = AB Let AB = BC = a Then by pythagoras theorem, \(AC^2\) = \(AB^2\) + \(BC^2\) = \(a^2\) + \(a^2\) …
Solution : The value of tan 45 degrees is 1. Proof : Let ABC be a triangle, right angled at B, in which \(\angle\) A = \(\angle\) C = 45 degrees \(\therefore\) BC = AB Let AB = BC = a Then by pythagoras theorem, \(AC^2\) = \(AB^2\) + \(BC^2\) = \(a^2\) + \(a^2\) = …
Solution : The value of cosec 60 degrees is \(2\over \sqrt{3}\). Proof : Consider an equilateral triangle ABC with each side of length of 2a. Each angle of \(\Delta\) ABC is of 60 degrees. Let AD be the perpendicular from A on BC. \(\therefore\) AD is the bisector of \(\angle\) A and D is the mid-point …
Solution : The value of sec 60 degrees is 2. Proof : Consider an equilateral triangle ABC with each side of length of 2a. Each angle of \(\Delta\) ABC is of 60 degrees. Let AD be the perpendicular from A on BC. \(\therefore\) AD is the bisector of \(\angle\) A and D is the mid-point …
Solution : The value of cot 60 degrees is \(1\over \sqrt{3}\). Proof : Consider an equilateral triangle ABC with each side of length of 2a. Each angle of \(\Delta\) ABC is of 60 degrees. Let AD be the perpendicular from A on BC. \(\therefore\) AD is the bisector of \(\angle\) A and D is the mid-point …
Solution : The value of tan 60 degrees is \(\sqrt{3}\). Proof : Consider an equilateral triangle ABC with each side of length of 2a. Each angle of \(\Delta\) ABC is of 60 degrees. Let AD be the perpendicular from A on BC. \(\therefore\) AD is the bisector of \(\angle\) A and D is the mid-point …
