Question

Anyone!! Topic =Trigonometry


Detailed solution please​

Answer 1

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

.Given that,

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

and

$\rm :\longmapsto\:cot \theta \: = \: \dfrac{5}{12}$

We know that,

$\rm :\longmapsto\: {cosec}^{2}\theta \: - \: {cot}^{2}\theta \: = 1$

$\rm :\longmapsto\: {cosec}^{2}\theta \: - \: {\bigg(\dfrac{5}{12} \bigg) }^{2} \: = 1$

$\rm :\longmapsto\: {cosec}^{2}\theta \: - \: {\bigg(\dfrac{25}{144} \bigg) } \: = 1$

$\rm :\longmapsto\: {cosec}^{2}\theta \: \: = \: 1 + \dfrac{25}{144}$

$\rm :\longmapsto\: {cosec}^{2}\theta \: \: = \: \dfrac{144 + 25}{144}$

$\rm :\longmapsto\: {cosec}^{2}\theta \: \: = \: \dfrac{169}{144}$

$\rm :\longmapsto\: {cosec}\theta \: \: = \: \pm \: \dfrac{13}{12}$

As,

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

$\rm :\implies\:cosec\theta \: = - \: \dfrac{13}{12}$

$\rm :\implies\:sin\theta \: = - \: \dfrac{12}{13}$

Also, we know that,

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

$\rm :\longmapsto\: {cos}^{2}\theta \: + \: {\bigg(\dfrac{ - 12}{13} \bigg) }^{2} \: = 1$

$\rm :\longmapsto\: {cos}^{2}\theta \: + \: {\bigg(\dfrac{ 144}{169} \bigg) } \: = 1$

$\rm :\longmapsto\: {cos}^{2}\theta \: \: = 1 - \dfrac{144}{169}$

$\rm :\longmapsto\: {cos}^{2}\theta \: \: = \dfrac{169 - 144}{169}$

$\rm :\longmapsto\: {cos}^{2}\theta \: \: = \dfrac{25}{169}$

$\rm :\longmapsto\: {cos}\theta \: = \: \pm \: \dfrac{5}{13}$

As,

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

$\rm :\implies\:cos\theta \: = - \: \dfrac{5}{13}$

Now,

Consider,

$\rm :\longmapsto\:2sin\theta \: + 3cos\theta \:$

$\rm \: = \: \: - 2 \times \dfrac{12}{13} - 3 \times \dfrac{5}{13}$

$\rm \: = \: \: - \dfrac{24}{13} - \dfrac{15}{13}$

$\rm \: = \: \: \dfrac{ - 24 - 15}{13}$

$\rm \: = \: \: \dfrac{ - 39}{13}$

$\rm \: = \: \: - 3$

$\bf\implies \:\:2sin\theta \: + 3cos\theta \: = - \: 3$

$\bf\implies \:Option \: (d) \: is \: correct$

Additional Information :-

Sign of Trigonometric ratios in Quadrants

sin (90°-θ)  =  cos θ

cos (90°-θ)  =  sin θ

tan (90°-θ)  =  cot θ

csc (90°-θ)  =  sec θ

sec (90°-θ)  =  csc θ

cot (90°-θ)  =  tan θ

sin (90°+θ)  =  cos θ

cos (90°+θ)  =  -sin θ

tan (90°+θ)  =  -cot θ

csc (90°+θ)  =  sec θ

sec (90°+θ)  =  -csc θ

cot (90°+θ)  =  -tan θ

sin (180°-θ)  =  sin θ

cos (180°-θ)  =  -cos θ

tan (180°-θ)  =  -tan θ

csc (180°-θ)  =  csc θ

sec (180°-θ)  =  -sec θ

cot (180°-θ)  =  -cot θ

sin (180°+θ)  =  -sin θ

cos (180°+θ)  =  -cos θ

tan (180°+θ)  =  tan θ

csc (180°+θ)  =  -csc θ

sec (180°+θ)  =  -sec θ

cot (180°+θ)  =  cot θ

sin (270°-θ)  =  -cos θ

cos (270°-θ)  =  -sin θ

tan (270°-θ)  =  cot θ

csc (270°-θ)  =  -sec θ

sec (270°-θ)  =  -csc θ

cot (270°-θ)  =  tan θ

sin (270°+θ)  =  -cos θ

cos (270°+θ)  =  sin θ

tan (270°+θ)  =  -cot θ

csc (270°+θ)  =  -sec θ

sec (270°+θ)  =  cos θ

cot (270°+θ)  =  -tan θ