1.tana & tanb are root of x²+px+q=0then value

sin²(a+b) + psin(a+b)cos(a+b) + qcos²(a+b)=

2.number of possible order pair (a,b) for which the inequality

a(cosx -1) + b² =cos(ax+b²) - 1 hold ture for all x belong to real no.

1. use sum of the roots and product of the roots to obtain the value of  tan(a+b).

after u get the value of tan(a+b) convert the given expression in terms of tan(a+b) and tan²(a+b) by taking common the cosine term.

put the value of tan(a+b) and get the answer!!!!

2.this problem can be solved easily using the graphs.

draw graphs for the expressions on the right hand side and the left hand side and see the portions in which they intersect.

u dont even need to sove the points where they intersect  you just count the intervals .Thats the answer.