Question
Question: If sin x + cosec x = 3, then find the value of \(\dfrac{1+{{\sin }^{4}}x}{{{\sin }^{2}}x}\)....
If sin x + cosec x = 3, then find the value of sin2x1+sin4x.
Solution
We will use the concept of cosecA=sinA1 to get the value of sin2x1+sin4x , cosecx is reciprocal of sinx. Then, we will try to get the expression of sin2x1+sin4x by solving and re -arranging the equation after the substation of cosecA=sinA1 . Also, we need to use formula of (a+b)2=a2+b2+2ab .
Complete step-by-step answer :
Now, we know that cosecA is reciprocal of sinA that is cosecA=sinA1 .
So, we can write sinx+cosecx=sinx+sinx1.
Then, sinx+sinx1=3 .
Taking L.C.M of sinx and sinx1,
sinxsin2x+1=3 .
Taking sinx in denominator on left hand side to the numerator of right hand side using cross multiplication,
sin2x+1=3×sinx ……( i )
Now, squaring both sides in equation ( i ), we get
sin4x+2sin2x+1=9sin2x , as (a+b)2=a2+b2+2ab
Taking 2sin2x from left hand side to right hand side, we get
sin4x+1=9sin2x−2sin2x,
On simplifying by subtracting 2sin2x from 9sin2x , we get
sin4x+1=7sin2x,
Taking sin2x in numerator on right hand side to the denominator of left hand side using cross multiplication ,
sin2x1+sin4x=7, which is equals to same expression mentioned in the question.
Hence, the answer we obtain is equals to 7 .
Note :You can also solve the equation sin2x+1=3×sinxby reducing it into quadratic equation by letting sinx = t which will give us quadratic equation t2−3t+1=0 which can further be solved for value of sin x and then we can substitute the obtained value in expression sin2x1+sin4x to get it is value. Always remember that cosecA=sinA1. Cross multiplication should be done in such a way that you get some expression which is given in question or it can be reduced to the same expression in question.