Question

if x is an acute angle what is the minimum value of tan^2x +cot^2x​

Answer 1

Minimum value = 2.

Step-by-step explanation:

$Let \: \: \: tan x = y, \: \: then \: \: cot x = \frac{1}{y}$

Then,

${tan}^{2} x \: + \: {cot}^{2} x \: = {y}^{2} + \frac{1}{ {y}^{2} }$

${y}^{2} + \frac{1}{ {y}^{2} } = {(y - \frac{1}{y} )}^{2} + 2$

Now, the minimum value of

${y}^{2} + \frac{1}{ {y}^{2} } =$

will occur when,

${(y - \frac{1}{y} )}^{2}$

will be minimum and minimum value of

${(y - \frac{1}{y} )}^{2} \: will \: \: be \: \: zero \: .$

Hence, the minimum value of

${y}^{2} + \frac{1}{ {y}^{2} } = {tan}^{2} x \: + {cot}^{2} x \: \: will \: \: be \: 2.$

Answer = 2.

Answer 2

Answer:

The minimum value is $2$.

Step-by-step explanation:

Assuming $tanx=y$ and $cot x=\frac{1}{y}$

Adding and squaring both equations

$tanx^{2} + cotx^{2} = y^{2} + (\frac{1}{y} )^{2}$

$y^{2} + (\frac{1}{y} )^{2}=(y-\frac{1}{2} )^{2} +2$

Minimum value of $y^{2} + (\frac{1}{y} )^{2}$ when $(y-\frac{1}{2} )^{2}$ is zero.

Hence, the required minimum value is $2$

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