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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 1+sin4xsin2x\dfrac{1+{{\sin }^{4}}x}{{{\sin }^{2}}x}.

Explanation

Solution

We will use the concept of cosecA=1sinA\cos ecA=\dfrac{1}{\sin A} to get the value of 1+sin4xsin2x\dfrac{1+{{\sin }^{4}}x}{{{\sin }^{2}}x} , cosecx\operatorname{cosec}x is reciprocal of sinxx. Then, we will try to get the expression of 1+sin4xsin2x\dfrac{1+{{\sin }^{4}}x}{{{\sin }^{2}}x} by solving and re -arranging the equation after the substation of cosecA=1sinA\cos ecA=\dfrac{1}{\sin A} . Also, we need to use formula of (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab .

Complete step-by-step answer :
Now, we know that cosecA\cos ecA is reciprocal of sinA\sin A that is cosecA=1sinA\cos ecA=\dfrac{1}{\sin A} .
So, we can write sinx+cosecx=sinx+1sinx\sin x+\cos ecx=\sin x+\dfrac{1}{\sin x}.
Then, sinx+1sinx=3\sin x+\dfrac{1}{\sin x}=3 .
Taking L.C.M of sinx\sin x and 1sinx\dfrac{1}{\sin x},
sin2x+1sinx=3\dfrac{{{\sin }^{2}}x+1}{\sin x}=3 .
Taking sinx\sin x in denominator on left hand side to the numerator of right hand side using cross multiplication,
sin2x+1=3×sinx{{\sin }^{2}}x+1=3\times \sin x ……( i )
Now, squaring both sides in equation ( i ), we get
sin4x+2sin2x+1=9sin2x{{\sin }^{4}}x+2{{\sin }^{2}}x+1=9{{\sin }^{2}}x , as (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab
Taking 2sin2x2{{\sin }^{2}}x from left hand side to right hand side, we get
sin4x+1=9sin2x2sin2x{{\sin }^{4}}x+1=9{{\sin }^{2}}x-2{{\sin }^{2}}x,
On simplifying by subtracting 2sin2x2{{\sin }^{2}}x from 9sin2x9{{\sin }^{2}}x , we get
sin4x+1=7sin2x{{\sin }^{4}}x+1=7{{\sin }^{2}}x,
Taking sin2x{{\sin }^{2}}x in numerator on right hand side to the denominator of left hand side using cross multiplication ,
1+sin4xsin2x=7\dfrac{1+{{\sin }^{4}}x}{{{\sin }^{2}}x}=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×sinx{{\sin }^{2}}x+1=3\times \sin xby reducing it into quadratic equation by letting sinx\sin x = t which will give us quadratic equation t23t+1=0{{t}^{2}}-3t+1=0 which can further be solved for value of sin x and then we can substitute the obtained value in expression 1+sin4xsin2x\dfrac{1+{{\sin }^{4}}x}{{{\sin }^{2}}x} to get it is value. Always remember that cosecA=1sinA\cos ecA=\dfrac{1}{\sin A}. 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.