 Hello and welcome to the session the given question says, the resultant of two forces p and q acting at an angle theta is equal to 2m plus 1 into root over p square plus q square and when they act at an angle pi by 2 minus theta the resultant is equal to 2m minus 1 into root over p square plus q square show that tan theta is equal to m minus 1 divided by n plus 1. Let's start with the solution now if the forces p q in angle theta then we are given that the resultant is 2m plus 1 into root over p square plus q square which implies that 2m plus 1 into root over p square plus q square is equal to root over p square plus q square plus 2 times of pq into cos theta. Since we know that if r is the magnitude of the resultant of two forces p and q acting as an angle alpha then r is equal to root over p square plus q square plus 2 pq cos theta. Here we have again two same forces p and q so we have root over p square plus q square plus 2 pq into cos theta as the resultant and we are given the resultant equal to 2m plus 1 into root over p square plus q square hence we have this equation. Now on squaring both the sides on the left hand side we have 2m plus 1 whole square into p square plus q square is equal to p square plus q square plus 2 times of pq cos theta which implies that 2m plus 1 whole square into p square plus q square minus p square plus q square taking both these terms on the left hand side on the right hand side we are left with 2pq cos theta which implies that 2m plus 1 square minus 1 into p square plus q square which is equal to 2pq cos theta or we have here applying the formula of a square minus b square which is a plus b a minus b we have 2m plus 1 minus 1 into 2m plus 1 plus 1 into p square plus q square is equal to 2pq cos theta or we have 2m into 2m plus 1 sorry 2m plus 2 into p square plus q square is equal to 2pq cos theta now dividing both the sides by 2 we have m 2 times of m into m plus 1 into p square plus q square is equal to p into q into cos theta here we have divided both the sides by 2 let this be equation number one now we are also given that f and q at an angle pi by 2 minus theta then the resultant is minus 1 into root over p square plus q square so we have minus 1 into root over p square plus q square is equal to root over p square plus q square plus 2pq into cos of pi by 2 minus theta and cos pi by 2 minus theta sine theta so next step can be written as 2m minus 1 into root over p square plus q square is equal to p square plus q square plus 2pq sine theta under the root or on a squaring both the sides we have 2m minus 1 whole square into p square plus q square is equal to p square plus q square plus 2pq sine theta this on a square in both the sides now taking p square plus q square on the left-hand side we have 2m minus 1 whole square into p square plus q square minus of p square plus q square is equal to 2pq sine theta of the further half taking p square plus q square common we have 2m minus 1 whole square minus 1 into p square plus q square and this is equal to 2pq sine theta now here find the identity a square minus p square since one can be written as one square as you can further written as 2m minus 1 minus 1 into 2m minus 1 plus 1 into p square plus q square is equal to 2pq sine theta or we have 2m minus 2 into 2m into p square plus q square is equal to 2pq sine theta or we have dividing both sides by 2 2m into m minus 1 into p square plus q square is equal to pq sine theta that was the equation number 2 now dividing equation number 2 by 1 the right-hand side of equation number 2 is pq sine theta and the right-hand side of equation 1 is pq cos theta so here we have pq cos theta similarly writing the left-hand side of equation number 2 we have 2m into m minus 1 into p square plus q square and the left-hand side of equation number 1 is 2m into m plus 1 into p square plus q square and on simplifying on the left-hand side we have sine theta divided by cos theta is equal to m minus 1 divided by m plus 1 which further implies that tan theta is equal to m minus 1 divided by m plus 1 this is what we are required to prove that tan theta is equal to m minus 1 divided by m plus 1 so this completes the session bye and take care