If a and b are non-zero positive real numbers and {cos{\ }}^2\theta=\frac{(a+b)^2}{4ab} (for all values

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Solution

Q.
If a and b are non-zero positive real numbers and ${c o s}^{2} θ = \frac{(a + b)^{2}}{4 a b}$ $(f o r a l l v a l u e s o f θ),$ then which of the following sets of values are true for a and b

(i) a = 1, b = 1 and a = 2, b = 2
(ii) a = 3, b = 3 and a = 6, b = 6
(iii) a = 5, b = 5 and a = 7, b = 7
(iv) a = 4, b = 4 and a = 8, b = 8

Choose the correct option from the following:

1 only (i)

2 (iii) and (iv)

3 (ii) and (iii)

4 All of them

Ans.
(4)

We have,

${c o s}^{2} θ = \frac{(a + b)^{2}}{4 a b} \Rightarrow 1 - {c o s}^{2} θ = 1 - \frac{(a + b)^{2}}{4 a b} {[T a k i n g n e g a t i v e a n d t h e n a d d i n g 1 b o t h t h e s i d e s] \Rightarrow {s i n}^{2} θ = \frac{4 a b - (a^{2} + b^{2} + 2 a b)}{4 a b} \Rightarrow {s i n}^{2} θ = \frac{4 a b - a^{2} - b^{2} - 2 a b}{4 a b} \Rightarrow {s i n}^{2} θ = \frac{- a^{2} - b^{2} + 2 a b}{4 a b} = - \frac{(a - b)^{2}}{4 a b}$

Since a, b are positive, the only non-negative value of $- \frac{(a - b)^{2}}{4 a b}$

is zero, which happens when a = b.

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