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

Solution : The value of Cos 60 degrees is \(1\over 2\). 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

Solution : The value of sin 60 degrees is \(\sqrt{3}\over 2\). 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