 Hi, and welcome to the session. Let's work out the following question. The question says, the particle is projected so as to have a range r on a horizontal plane through the point of projection if alpha and beta are the possible angle of projections. And T1 and T2 are corresponding time-blow flight. Show that T1 squared minus T2 squared divided by T1 squared plus T2 squared is equal to sin alpha minus beta divided by sin alpha plus beta. Let us see the solution to this question. Let T1 and T2 be the times of flight when a particle is projected with velocity u and angle of projection v alpha and beta respectively. Therefore, T1 is equal to 2u sin alpha divided by g T2 is equal to 2u sin beta divided by g where g is acceleration due to gravity. Therefore, T1 squared upon T2 squared is equal to sin square alpha divided by sin square beta. Now, applying dividendo and componento, we get T1 squared minus T2 squared divided by T1 squared plus T2 squared is equal to sin square alpha minus sin square beta divided by sin square alpha plus sin square beta. This we call 1. Now, in both the flights, the horizontal range is the same. Therefore, r is equal to u square sin 2 alpha divided by g that is equal to u square sin 2 beta divided by g. This implies sin 2 alpha minus sin 2 beta is equal to 0. This implies 2 cos alpha plus beta sin alpha minus beta is equal to 0. This implies cos alpha minus, this is called alpha plus beta is equal to 0 or sin alpha minus beta is equal to 0. Now, this implies that alpha plus beta is equal to pi by 2. And this implies that alpha minus beta is equal to 0. But we see that since alpha is not equal to beta, therefore, alpha plus beta is equal to pi by 2. And this we call 2. This implies that alpha is equal to pi by 2 minus beta. Or we can say beta is equal to pi by 2 minus alpha. Now, t1 square minus t2 square divided by t1 square plus t2 square is equal to sin alpha into sin alpha that is sin square alpha minus sin beta into sin beta that is sin square beta divided by sin alpha into sin alpha plus sin beta into sin beta. This can be written further as sin alpha into sin pi by 2 minus beta minus sin beta into sin pi by 2 minus alpha divided by sin alpha into sin pi by 2 minus beta plus sin beta into sin pi by 2 minus alpha which is further equal to sin alpha cos beta minus sin beta cos alpha divided by sin alpha cos beta plus sin beta cos alpha. And this is further equal to sin alpha minus beta divided by sin alpha plus beta. So we have proved that t1 square minus t2 square divided by t1 square plus t2 square is equal to sin alpha minus beta divided by sin alpha plus beta. So this is what we were supposed to prove in this question. I hope that you understood the solution and enjoyed the session. Have a good day.