Question

tanx + 1/ tanx =2 ,then the value of tan^2 x+ 1/tan^2 x is equal to​

Answer 1

Question :-

$if \: \: tan \: x + \frac{1}{tan \: x} = 2 \: \: \: then \: find \: the \\ value \: of \: \red{{tan}^{2} x + \frac{1}{ {tan}^{2} x} }$

Answer :-

$\implies \: \boxed{ {tan}^{2} x \: + \frac{1}{ {tan}^{2} x} \: = 2} \:$

Formula used :-

$\implies \boxed{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}$

Step - by - step explanation :-

Given that,

$\implies \: tan \: x \: + \frac{1}{tan \: x} \: = 2$

Squaring on both sides,

${ \bigg(tan \: x \: + \frac{1}{tan \: x} \bigg)}^{2} = {2}^{2} \\ \\ \: \text{applied \: the \: given \: formula } \\ \\ \rightarrow \: \small {tan}^{2} x \: + \frac{1}{ {tan}^{2} x} + 2 \: tan \: x \: \frac{1}{tan \: x} = 4 \\ \\ \rightarrow \: {tan}^{2} x \: + \frac{1}{ {tan}^{2} x} \: + 2 = 4 \\ \\ \rightarrow \: {tan}^{2} x \: + \frac{1}{ {tan}^{2} x} \: = 4 - 2 \\ \\ \rightarrow \: \red{\boxed{ {tan}^{2} x \: + \frac{1}{ {tan}^{2} x} \: = 2}}$

This is the required solution.

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