Select Page

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.

## Problem

If $\displaystyle { \frac{\sin^4 x }{a} + \frac{\cos^4 x }{b} = \frac{1}{a+b} }$ , then show that $\displaystyle { \frac{\sin^6 x }{a^2} + \frac{\cos^6 x }{b^2} = \frac{1}{(a+b)^2} }$

## Solution

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)

## Chatuspathi:

• What is this topic: Trigonometry
• What are some of the associated concept: Change of Variables
• Where can learn these topics: Cheenta I.S.I. & C.M.I. course, discusses these topics in the ‘Trigonometry’ module.
• Book Suggestions: Trigonometry Volume I by S.L. Loney