Question

4) If tan20 = 3 then find the value of seco​

Answer 1

Answer:

given tan^2O =3

square rooting on both sides we get

tanO=√3

when O = 60° we get tanO=√3

secO=sec60°=2

secO=2

or

given

tan^2O=3

we know that

sec^2a-tan^2a=1

substituting the given we get,

sec^2O-3=1

sec^2O=1+3

sec^2O=4

square rooting on both sides we get

secO=√4=2

secO=2

Answer 2

Appropriate :-

$\rm \: If \: tan2x = 3, \: find \: the \: value \: of \: secx$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: tan2x = 3$

$\rm \: \dfrac{2 \: tanx}{1 - {tan}^{2}x} = 3$

$\rm \: 3 - 3 {tan}^{2}x = 2tanx$

$\rm \: 3{tan}^{2}x + 2tanx - 3 = 0$

Its a quadratic in tanx, so roots are evaluated by using Quadratic formula. So,

$\rm \: tanx = \dfrac{ - 2 \: \pm \: \sqrt{ {2}^{2} + 4 \times \times 3} }{2(3)}$

$\rm \: tanx = \dfrac{ - 2 \: \pm \: \sqrt{4 + 36} }{6}$

$\rm \: tanx = \dfrac{ - 2 \: \pm \: \sqrt{40} }{6}$

$\rm \: tanx = \dfrac{ - 2 \: \pm \: 2 \sqrt{10} }{6}$

$\rm \: tanx = \dfrac{ - 1 \: \pm \: \sqrt{10} }{3}$

$\rm \: tanx = \dfrac{ - 1+\sqrt{10} }{3} \: \: or \: \: tanx = \dfrac{ - 1 - \sqrt{10} }{3}$

Case :- 1

$\rm \: tanx = \dfrac{ - 1+\sqrt{10} }{3} \:$

$\rm\implies \:tanx > 0$

$\rm\implies \:x \: \in \: \bigg(0,\dfrac{\pi}{2} \bigg) \cup \: \bigg(\pi ,\dfrac{3\pi}{2} \bigg)$

$\rm\implies \:secx > 0 \: \: or \: \: secx < 0$

We know,

$\rm \: {sec}^{2}x - {tan}^{2}x = 1$

$\rm \: {sec}^{2}x = {tan}^{2}x + 1$

$\rm \: {sec}^{2}x = {\bigg[\dfrac{ \sqrt{10} - 1}{3} \bigg]}^{2} + 1$

$\rm \: {sec}^{2}x = \dfrac{10 + 1 - 2 \sqrt{10} }{9} + 1$

$\rm \: {sec}^{2}x = \dfrac{11 - 2 \sqrt{10} }{9} + 1$

$\rm \: {sec}^{2}x = \dfrac{11 - 2 \sqrt{10} + 9}{9}$

$\rm \: {sec}^{2}x = \dfrac{20 - 2 \sqrt{10}}{9}$

$\rm \: {sec}x = \: \pm \: \sqrt{\dfrac{20 - 2 \sqrt{10}}{9}}$

or

$\rm \: {sec}x = \: \pm \: \frac{ \sqrt{20 - 2 \sqrt{10} } }{3}$

Case :- 2

$\rm \:When \: tanx = \dfrac{ - 1 - \sqrt{10} }{3} \:$

$\rm\implies \:tanx < 0$

$\rm\implies \:x \: \in \: \bigg(\dfrac{\pi}{2} ,\pi\bigg) \: \cup \: \bigg(\dfrac{3\pi}{2} ,2\pi \bigg)$

$\rm\implies \:secx < 0 \: \: or \: \: secx > 0$

Now, we know

$\rm \: {sec}^{2}x - {tan}^{2}x = 1$

$\rm \: {sec}^{2}x = {tan}^{2}x + 1$

$\rm \: {sec}^{2}x = {\bigg[\dfrac{ - \sqrt{10} - 1}{3} \bigg]}^{2} + 1$

$\rm \: {sec}^{2}x = {\bigg[\dfrac{ \sqrt{10} + 1}{3} \bigg]}^{2} + 1$

$\rm \: {sec}^{2}x = \dfrac{10 + 1 + 2 \sqrt{10} }{9} + 1$

$\rm \: {sec}^{2}x = \dfrac{11 + 2 \sqrt{10} }{9} + 1$

$\rm \: {sec}^{2}x = \dfrac{11 + 2 \sqrt{10} + 9}{9}$

$\rm \: {sec}^{2}x = \dfrac{20 + 2 \sqrt{10}}{9}$

$\rm \: {sec}x = \: \pm \: \sqrt{\dfrac{20 + 2 \sqrt{10}}{9}}$

or

$\rm \: {sec}x = \: \pm \: \frac{ \sqrt{20 + 2 \sqrt{10} } }{3}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered} {\boxed{ \begin{array}{c} \underline{\underline{ \text{Additional \: lnformation}}} \\ \\ & \rm \: sin2x \: = 2 \: sinx \: cosx\:\\ & \rm \: cos2x = 1 - {2sin}^{2}x \\ & \rm \: cos2x = {2cos}^{2}x - 1 \\ & \rm \: cos2x = {cos}^{2}x - {sin}^{2}x \\ & \rm \:tan2x = \dfrac{2tanx}{1 - {tan}^{2} x} \\ & \rm \: sin2x = \dfrac{2tanx}{1 + {tan}^{2}x } \\ & \rm \:sin3x = 3sinx - {4sin}^{3}x \\ & \rm \: cos3x = {4cos}^{3}x - 3cosx \\ & \rm \: tan3x = \dfrac{3tanx - {tan}^{3} x}{1 - {3tan}^{2}x} \end{array}}}\end{gathered}$