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1.tana & tanb are root of x 2 +px+q=0then value sin 2 (a+b) + psin(a+b)cos(a+b) + qcos 2 (a+b)= 2.number of possible order pair (a,b) for which the inequality a(cosx -1) + b 2 =cos(ax+b 2 ) - 1 hold ture for all x belong to real no.

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.


 

Grade:12

1 Answers

AskIITianExpert Srijan.iitd
8 Points
12 years ago

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.

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