Solution : Draw a right angled triangle ABC right angled at B. Let \(\angle\) A = \(\theta\), then \(\angle\) C = 90 – \(\theta\) cos A = \(cos\theta\) = \(AB\over AC\) ……..(1) sin C = \(sin(90 – \theta)\) = \(AB\over AC\) …
Prove that \(Sin(90 – \theta)\) = \(Cos\theta\). Read More »
Solution : The value of cosec 90 degrees is 1. Proof : \(\angle\) A of \(\Delta\) ABC is made large and large until it becomes 90 degrees. As \(\angle\) A gets large and large \(\angle\) C gets smaller and smaller. Side AB goes on decreasing. The point A gets closer to point B. When \(\angle\) …
Solution : The value of sec 90 degrees is \(\infty\) or not defined. Proof : \(\angle\) A of \(\Delta\) ABC is made large and large until it becomes 90 degrees. As \(\angle\) A gets large and large \(\angle\) C gets smaller and smaller. Side AB goes on decreasing. The point A gets closer to point …
Solution : The value of tan 90 degrees is \(\infty\) or not defined. Proof : \(\angle\) A of \(\Delta\) ABC is made large and large until it becomes 90 degrees. As \(\angle\) A gets large and large \(\angle\) C gets smaller and smaller. Side AB goes on decreasing. The point A gets closer to point B. …
Solution : The value of cot 90 degrees is 0. Proof : \(\angle\) A of \(\Delta\) ABC is made large and large until it becomes 90 degrees. As \(\angle\) A gets large and large \(\angle\) C gets smaller and smaller. Side AB goes on decreasing. The point A gets closer to point B. When \(\angle\) …
Solution : The value of cos 90 degrees is 0. Proof : \(\angle\) A of \(\Delta\) ABC is made large and large until it becomes 90 degrees. As \(\angle\) A gets large and large \(\angle\) C gets smaller and smaller. Side AB goes on decreasing. The point A gets closer to point B. When \(\angle\) …
Solution : The value of sin 90 degrees is 1. Proof : \(\angle\) A of \(\Delta\) ABC is made large and large until it becomes 90 degrees. As \(\angle\) A gets large and large \(\angle\) C gets smaller and smaller. Side AB goes on decreasing. The point A gets closer to point B. When \(\angle\) …
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, \(AC^2\) = \(AB^2\) + \(BC^2\) = \(a^2\) + \(a^2\) = …
