Question

If tan alpha+1/tan alpha=2 find the value of tan^2alpha+1/tan^2alpha

Answer 1

Answer:

answer for the given problem is given in both methods

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{tan^{2}\:\alpha+\frac{1}{tan^{2}\:\alpha}=2}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies tan \: \alpha + \frac{1}{tan \: \alpha } = 2 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies {tan}^{2} \: \alpha + \frac{1}{ {tan}^{2} \: \alpha } = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies tan \: \alpha + \frac{1}{tan \: \alpha } = 2 \\ \\ \text{Squaring \: both \: side } \\ \tt: \implies \bigg({tan \: \alpha + \frac{1}{ tan \: \alpha } } \bigg)^{2} = {2}^{2} \\ \\ \tt \circ \: ( {a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ \\ \tt: \implies {tan}^{2} \: \alpha + \frac{1}{ {tan}^{2} \: \alpha } + 2 \times tan \: \alpha \times \frac{1}{tan \: \alpha } = 4 \\ \\ \tt: \implies {tan}^{2} \: \alpha + \frac{1}{ {tan}^{2} \: \alpha } + 2 = 4 \\ \\ \tt: \implies {tan}^{2} \: \alpha + \frac{1}{ {tan}^{2} \: \alpha } = 4 - 2 \\ \\ \green{\tt: \implies {tan}^{2} \: \alpha + \frac{1}{ {tan}^{2} \: \alpha } =2}$