Question

Find the possible values of tan x, if cos²x + 5 sin x in to
cos x = 3.​

Answer 1

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

$\sf \: {cos}^{2} x \: + \: 5sinx \: cosx \: = 3$

$\large\underline{\sf{To\:Find - }}$

$\sf \: value \: of \: tanx$

$\large\underline{\sf{Understanding \: the \: concept - }}$

Since, we have to find the value of tanx, so we have firstly convert the given equation in terms of tanx by dividing the whole equation by cos²x. Then, we factorize the given equation using splitting of middle terms.

Splitting of middle terms :-

• In order to factorize  ax² + bx + c we have to find numbers p and q such that p + q = b and pq = ac.

• After finding p and q, we split the middle term in the quadratic as px + qx and get desired factors by grouping the terms.

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$1. \: \: \: \boxed{ \bf{tanx \: = \: \dfrac{sinx}{cosx} }}$

$2. \: \: \: \boxed{ \bf{secx \: = \: \dfrac{1}{cosx} }}$

$3. \: \: \: \boxed{ \bf{ {sec}^{2} x - {tan}^{2} x \: = \: 1}}$

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

Given that,

$\rm :\longmapsto\: {cos}^{2} x \: + \: 5 \: sinx \: cosx \: = \: 3$

On dividing both sides by cos²x, we get

$\rm :\longmapsto\:\dfrac{ {cos}^{2} x}{ {cos}^{2}x } + \dfrac{5sinx \: cosx}{ {cos}^{2}x } = \dfrac{3}{ {cos}^{2}x }$

$\rm :\longmapsto\:1 \: + \: 5 \: tanx \: = \: 3 \: {sec}^{2} x$

$\rm :\longmapsto\:1 \: + \: 5 \: tanx \: = \: 3 \:(1 + {tan}^{2} x)$

$\rm :\longmapsto\:1 \: + \: 5 \: tanx \: = 3 + \: 3 \: {tan}^{2} x$

$\rm :\longmapsto\:\: 3 \: {tan}^{2} x \: - \: 5tanx \: + \: 2 \: = \: 0$

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

$\rm :\implies\:3tanx(tanx - 1) - 2(tanx - 1) = 0$

$\rm :\longmapsto\:(tanx - 1)(3tanx - 2) = 0$

$\rm :\longmapsto\:tanx - 1 = 0 \: \: or \: \: 3tanx - 2 = 0$

$\bf\implies \:tanx = 1 \: \: \: or \: \: \: tanx = \dfrac{2}{3}$

Additional Information :-

$1. \: \: \: \boxed{ \bf{cosecx \: = \: \dfrac{1}{sinx} }}$

$2. \: \: \: \boxed{ \bf{cotx \: = \: \dfrac{cosx}{sinx} }}$

$3. \: \: \: \boxed{ \bf{ {sin}^{2}x + {cos}^{2}x = 1}}$

$4. \: \: \: \boxed{ \bf{ {cosec}^{2} x - {cot}^{2} x \: = \: 1}}$