Question

solve this
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Answer 1

Answer:

Step-by-step explanation:

Let us suppose

$\sf{\beta=\dfrac{a^2}{cos^2(x)}+\dfrac{b^2}{sin^2(x)}}$

$\sf{\implies\beta=a^2\,sec^2(x)+b^2\,cosec^2(x)}$

$\sf{\implies\beta=a^2\left(1+tan^2(x)\right)+b^2\left(1+cot^2(x)\right)}$

$\sf{\implies\beta=a^2+a^2\,tan^2(x)+b^2+b^2\,cot^2(x)}$

$\sf{\implies\beta=a^2\,tan^2(x)+b^2\,cot^2(x)-2ab+a^2+b^2+2ab}$

$\sf{\implies\beta=a^2\,tan^2(x)+b^2\,cot^2(x)-2\cdot\,a\,tan(x)\cdot\,b\,cot(x)+\left(a+b\right)^2}$

$\sf{\implies\beta=\left\{a\,tan(x)-b\,cot(x)\right\}^2+\left(a+b\right)^2}$

$\sf{\implies\beta=\left\{a\,tan(x)-\dfrac{b}{tan(x)}\right\}^2+\left(a+b\right)^2}$

Now,

We know,

$\sf{\left\{a\,tan(x)-\dfrac{b}{tan(x)}\right\}^2\ge0}$

$\sf{\implies\left\{a\,tan(x)-\dfrac{b}{tan(x)}\right\}^2+\left(a+b\right)^2\ge0+\left(a+b\right)^2}$

$\sf{\implies\beta\ge\left(a+b\right)^2}$

Hence, the minimum value of the given expression is (a+b)²

Answer 2

hm gd kaal se m on nie aaungi..