This is a Test of Mathematics Solution Subjective 115 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
If $\displaystyle { \frac{\sin^4 x }{a} + \frac{\cos^4 x }{b} = \frac{1}{a+b} }$ , then show that $ {\frac{1}{(a+b)^2} }$
Put $ \sin^2 x = t $.
The given expression reduces to $ \displaystyle { \frac{t^2 }{a} + \frac{1+t^2 -2t }{b} = \frac{1}{a+b} }$
$ \displaystyle {\Rightarrow (a+b)t^2 -2at +a - \frac{ab}{a+b} = 0 }$
$ \displaystyle {\Rightarrow (a+b)^2t^2 -2a(a+b)t +a^2 = 0 }$
$ \displaystyle {\Rightarrow ((a+b)t-a)^2=0 }$
$ \displaystyle {\Rightarrow \frac{a}{a+b}=t }$
$ \displaystyle {\Rightarrow \frac{b}{a+b}=1-t }$
Hence replacing in the required expression we get
$ \displaystyle { \frac{\sin^6 x }{a^2} + \frac{\cos^6 x }{b^2} }$
$ \displaystyle {=\frac{t^3 }{a^2} + \frac{(1-t)^3 }{b^2}}$
$ \displaystyle {=\frac{a^3 }{(a+b)^3a^2} + \frac{b^3 }{(a+b)^3b^2}}$
$ \displaystyle {=\frac{a}{(a+b)^3} + \frac{b}{(a+b)^3} }$
$ \displaystyle {=\frac{a+b}{(a+b)^3} }$
$ \displaystyle {=\frac{1}{(a+b)^2} }$
(Proved)
This is a Test of Mathematics Solution Subjective 115 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
If $\displaystyle { \frac{\sin^4 x }{a} + \frac{\cos^4 x }{b} = \frac{1}{a+b} }$ , then show that $ {\frac{1}{(a+b)^2} }$
Put $ \sin^2 x = t $.
The given expression reduces to $ \displaystyle { \frac{t^2 }{a} + \frac{1+t^2 -2t }{b} = \frac{1}{a+b} }$
$ \displaystyle {\Rightarrow (a+b)t^2 -2at +a - \frac{ab}{a+b} = 0 }$
$ \displaystyle {\Rightarrow (a+b)^2t^2 -2a(a+b)t +a^2 = 0 }$
$ \displaystyle {\Rightarrow ((a+b)t-a)^2=0 }$
$ \displaystyle {\Rightarrow \frac{a}{a+b}=t }$
$ \displaystyle {\Rightarrow \frac{b}{a+b}=1-t }$
Hence replacing in the required expression we get
$ \displaystyle { \frac{\sin^6 x }{a^2} + \frac{\cos^6 x }{b^2} }$
$ \displaystyle {=\frac{t^3 }{a^2} + \frac{(1-t)^3 }{b^2}}$
$ \displaystyle {=\frac{a^3 }{(a+b)^3a^2} + \frac{b^3 }{(a+b)^3b^2}}$
$ \displaystyle {=\frac{a}{(a+b)^3} + \frac{b}{(a+b)^3} }$
$ \displaystyle {=\frac{a+b}{(a+b)^3} }$
$ \displaystyle {=\frac{1}{(a+b)^2} }$
(Proved)
