Question

least the value of 25cosec^2x+36sec^2x is
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Answer 1

To find:

least the value of $25cosec^2x+36sec^2x$ is ​

Solution:

Let us have a look at a few formula:

$1.\ cosec^2x = 1+ cot^2x\\2. \sec^2x = 1+ tan^2x\\3. \ tanx = \dfrac{1}{cotx}$

Let $y$ = $25cosec^2x+36sec^2x$

Using 1 and 2 in y:

$y=25(1+cot^2x)+36(1+tan^2x)\\\Rightarrow y=25+25cot^2x+36+36tan^2x\\\Rightarrow y=61+25cot^2x+36tan^2x\\\Rightarrow y=61+25cot^2x+36\frac{1}{cot^2x}$

For cotx terms, let us use the following formula:

Arithmetic mean $\geq$ Geometric mean

$\dfrac{25cot^2x+36\frac{1}{cot^2x}}{2} \geq \sqrt{25cot^2x\times 36\frac{1}{cot^2x}}\\\Rightarrow \dfrac{25cot^2x+36\frac{1}{cot^2x}}{2} \geq \sqrt{25\times 36}\\\Rightarrow 25cot^2x+36\frac{1}{cot^2x} \geq 30 \times 2\\\Rightarrow 25cot^2x+36\frac{1}{cot^2x} \geq 60$

Putting the above value in $y$:

So $y_{min} = 61+60 = \bold{121}$

Therefore, the minimum value of $y$ = $25cosec^2x+36sec^2x$ is 121.