Question

If 1+ sin^2 theta =3 sin theta x cos theta, then tan theta=​

Answer 1

Step-by-step explanation:

solve the cubic and u wil get the value of tan theta

Answer 2

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

$\rm :\longmapsto\:1 + {sin}^{2}\theta = 3sin\theta \: cos\theta$

$\red{\rm :\longmapsto\:Divide \: both \: sides \: by \: {cos}^{2}\theta}$

$\rm :\longmapsto\:\dfrac{1}{ {cos}^{2}\theta} + \dfrac{ {sin}^{2}\theta}{ {cos}^{2} \theta} = \dfrac{3sin\theta \: cos\theta}{ {cos}^{2} \theta}$

$\rm :\longmapsto\: {sec}^{2}\theta + {tan}^{2}\theta = 3tan\theta$

${\bigg \{ \because \: \sf \: \dfrac{1}{cos\theta} = sec\theta \bigg \}} \\ {\bigg \{ \because \: \sf \: \dfrac{sin\theta}{cos\theta} = tan\theta \bigg \}}$

$\rm :\longmapsto\: 1 + {tan}^{2}\theta + {tan}^{2}\theta = 3tan\theta$

${\bigg \{ \because \: \sf \: {sec}^{2}\theta - {tan}^{2}\theta = 1 \bigg \}}$

$\rm :\longmapsto\: 1 + 2{tan}^{2}\theta = 3tan\theta$

$\rm :\longmapsto\: {2tan}^{2}\theta - 3tan\theta + 1 = 0$

$\rm :\longmapsto\: {2tan}^{2}\theta - 2tan\theta - tan\theta + 1 = 0$

$\rm :\longmapsto\:2tan\theta(tan\theta - 1) - 1(tan\theta - 1) = 0$

$\rm :\longmapsto\:(tan\theta - 1)(2tan\theta - 1) = 0$

$\bf\implies \:tan\theta = 1 \: \: \: or \: \: \: tan\theta = \dfrac{1}{2}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identitiest

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1