 Hello and welcome to the session. Let us discuss the following question. It says find perpendicular distance from the origin of the line joining the points cos theta, sin theta and cos phi, sin phi. To solve this question we need to know the equation of line joining two points, say x1, y1 and x2, y2. It is given by y minus y1 is equal to y2 minus y1 upon x2 minus x1 into x minus x1. We also need to know the perpendicular distance of a point, say x1, y1 to the line given by ax plus by plus c is equal to 0. The distance is given by mod of ax1 plus by 1 plus c upon under root of a squared plus b squared. So, this knowledge will work as k idea for this question. Let us now move on to the solution. We first find equation of line joining cos theta, sin theta and cos phi, sin phi. It is given by y minus y1 where y1 is sin theta is equal to y2 minus y1, where y2 is sin phi minus y1 which is sin theta upon x2 minus x1, where x2 is cos phi and x1 is cos theta into x minus x1, where x1 is cos theta. Now this implies y minus sin theta is equal to sin phi minus sin theta is equal to 2 cos theta plus phi upon 2 into sin theta minus phi upon 2 upon cos phi minus cos theta and cos phi minus cos theta is equal to minus 2 sin theta plus phi upon 2 sin theta minus phi upon 2 into x minus cos theta. Now 2 gets cancelled with 2 and sin theta minus phi by 2 gets cancelled with sin theta minus phi by 2 and this implies y minus sin theta is equal to minus cos theta plus phi upon 2 upon sin theta plus phi upon 2 into x minus cos theta and this again implies y minus sin theta into sin theta plus phi by 2 is equal to minus cos theta plus phi by 2 into x minus cos theta and this again implies sin theta plus phi upon 2 into y is sin theta plus y upon 2 y minus sin theta into sin theta plus phi upon 2 is equal to minus x cos theta plus phi upon 2 minus minus cos theta into cos theta plus phi upon 2. Now this implies transposing minus x cos theta plus phi by 2 to the left hand side we have x cos theta plus phi upon 2 plus sin theta plus phi upon 2 into y is equal to cos theta plus phi upon 2 into cos theta we also transpose minus sin theta sin theta plus phi by 2 to the right hand side so that it becomes plus sin theta sin theta plus phi by 2 and this implies x cos theta plus phi by 2 plus y sin theta plus phi by 2 is equal to cos theta cos theta plus phi by 2 plus sin theta sin theta plus phi by 2. Now to the RHS we apply the formula of cos A minus B which is equal to cos A cos B plus sin A sin B here A is theta and B is theta plus phi by 2 so this implies x cos theta plus phi by 2 plus y sin theta plus phi by 2 is equal to cos theta minus theta plus phi by 2 and this implies x cos theta plus phi by 2 plus y sin theta plus phi by 2 is equal to cos theta minus phi by 2 and this is the equation of the line joining cos theta sin theta cos phi sin phi. Now we find perpendicular distance from origin that is the point 0 0 to the line joining cos theta sin theta cos phi sin phi and this is denoted by 1. Distance of a point to the line is given by this formula that is A x1 plus B y1 plus C upon under root of A square plus B square so the perpendicular distance from 0 0 to the line is given by mod of 0 into cos theta plus phi upon 2 plus 0 into sin theta plus phi upon 2 minus cos theta minus phi upon 2 upon under root of square of coefficient of x that is cos square theta plus phi upon 2 plus square of coefficient of y that is sin square theta plus phi upon 2 and now this is equal to mod of cos theta minus phi upon 2 since the denominator is 1 as cos square theta plus phi upon 2 plus sin square theta plus phi upon 2 is 1. Again this is equal to mod of cos phi minus theta upon 2 as we know that cos of minus x is equal to cos of x so here we took minus common and then we got cos phi minus theta upon 2. Now we multiply and divide by 2 sin phi minus theta upon 2 so we have mod of 2 sin phi minus theta upon 2 cos phi minus theta upon 2 upon sin phi minus theta upon 2 and now this is equal to mod of sin phi minus theta upon 2 upon sin phi minus theta upon 2 this is by applying the formula for sin 2 theta which is equal to 2 sin theta cos theta now this is equal to mod of sin phi minus theta upon mod 2 sin phi minus theta upon 2 which is again equal to mod of sin phi minus theta upon 2 into mod sin phi minus theta upon 2 hence the perpendicular distance from the origin is mod sin phi minus theta upon 2 into mod sin phi minus theta upon 2 so this completes the question hope you enjoy the session goodbye and take care