Question

if
$\tan( a ) = \sqrt{3}$
and 'a' lies in the third quadrant, then what is sin(a) + cos(a)
​

Answer 1

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

Given that .

$\rm :\longmapsto\:\pi < a < \dfrac{3\pi}{2}$

and

$\rm :\longmapsto\:tana \: = \sqrt{3}$

We know that

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

On substituting the value of tan a, we get

$\rm :\longmapsto\: {sec}^{2}a - {( \sqrt{3} )}^{2} = 1$

$\rm :\longmapsto\: {sec}^{2}a - 3 = 1$

$\rm :\longmapsto\: {sec}^{2}a = 1 + 3$

$\rm :\longmapsto\: {sec}^{2}a = 4$

$\rm :\longmapsto\: {sec}a = \: \pm \: 2$

As,

$\rm :\longmapsto\:\pi < a < \dfrac{3\pi}{2}$

So, seca < 0

$\bf\implies \:seca = - 2$

So,

$\rm :\longmapsto\:cosa = - \: \dfrac{1}{2}$

Now,

We know that

$\rm :\longmapsto\: {sin}^{2}a + {cos}^{2}a = 1$

$\rm :\longmapsto\: {sin}^{2}a + {\bigg( - \dfrac{1}{2} \bigg) }^{2} = 1$

$\rm :\longmapsto\: {sin}^{2}a + {\bigg( \dfrac{1}{4} \bigg) } = 1$

$\rm :\longmapsto\: {sin}^{2}a = 1 - \dfrac{1}{4}$

$\rm :\longmapsto\: {sin}^{2}a = \dfrac{4 - 1}{4}$

$\rm :\longmapsto\: {sin}^{2}a = \dfrac{3}{4}$

$\rm :\longmapsto\: {sin}a = \: \pm \: \dfrac{ \sqrt{3} }{2}$

As,

$\rm :\longmapsto\:\pi < a < \dfrac{3\pi}{2}$

So,

$\rm :\longmapsto\: {sin}a = \: - \: \dfrac{ \sqrt{3} }{2}$

Now,

Consider,

$\rm :\longmapsto\:sina + cosa$

$\rm \: = \: - \dfrac{ \sqrt{3} }{2} - \dfrac{1}{2}$

$\rm \: = \: - \: \dfrac{ \sqrt{3} + 1}{2}$

Hence,

$\bf\implies \:sina + cosa= \: - \: \dfrac{ \sqrt{3} + 1}{2}$

Alternative Method :-

Given that,

$\rm :\longmapsto\:\pi < a < \dfrac{3\pi}{2}$

and

$\rm :\longmapsto\:tana \: = \sqrt{3}$

$\rm :\longmapsto\:tana \: = tan\bigg(\dfrac{\pi}{3} \bigg)$

$\rm :\longmapsto\:tana \: = tan\bigg(\pi + \dfrac{\pi}{3} \bigg)$

$\rm :\longmapsto\:tana \: = tan\bigg(\dfrac{3\pi + \pi}{3} \bigg)$

$\rm :\longmapsto\:tana \: = tan\bigg(\dfrac{4\pi}{3} \bigg)$

$\bf\implies \:a = \dfrac{4\pi}{3}$

Now, Consider

$\rm :\longmapsto\:sina + cosa$

$\rm \: = \: sin \dfrac{4\pi}{3} + cos \dfrac{4\pi}{3}$

$\rm \: = \: sin\bigg(\pi + \dfrac{\pi}{3} \bigg) + cos\bigg(\pi + \dfrac{\pi}{3} \bigg)$

$\rm \: = \: - sin \dfrac{\pi}{3} - cos \dfrac{\pi}{3}$

$\rm \: = \: - \dfrac{ \sqrt{3} }{2} - \dfrac{1}{2}$

$\rm \: = \: - \: \dfrac{ \sqrt{3} + 1}{2}$

Hence,

$\bf\implies \:sina + cosa= \: - \: \dfrac{ \sqrt{3} + 1}{2}$