Question

If tan^2 theta-5tan theta +1=0,then find the value of tan^2+ cot^2

Answer 1

$\textbf{Given:}$

$tan^2\theta-5\,tan\,\theta+1=0$

$\textbf{To find:}$

$\text{The value of tan^2\theta+cot^2\theta }$

$\textbf{Solution:}$

$\text{Consider,}$

$tan^2\theta+cot^2\theta$

$\text{Using the identity,}$

$\boxed{\bf\,a^2+b^2=(a+b)^2-2\,ab}$

$=(tan\theta+cot\theta)^2-2\,tan\theta\,cot\theta$

$=(tan\theta+\dfrac{1}{tan\theta})^2-2\,tan\theta(\dfrac{1}{tan\theta})$

$=(\dfrac{tan^2\theta+1}{tan\theta})^2-2$

$\text{Using,}\;\;\bf\,tan^2\theta+1=5\,tan\theta$

$=(\dfrac{5\,tan\theta}{tan\theta})^2-2$

$=(\dfrac{5}{1})^2-2$

$=25-2$

$=23$

$\textbf{Answer:}$

$\textbf{The value of \bf\,tan^2\theta+cot^2\theta is 23}$

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