Question

If 4cos thetha +3sin thetha =5 find value of tan thetha

Answer 1

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

Given Trigonometric equation is

$\rm :\longmapsto\:4cos\theta + sin\theta = 5$

$\boxed{ \bf{ \:Divide \: whole \: equation \: by \: cos\theta, \: we \: get}}$

$\rm :\longmapsto\:\dfrac{4cos\theta}{cos\theta} + \dfrac{3sin\theta}{cos\theta} = \dfrac{5}{cos\theta}$

$\rm :\longmapsto\:4 + 3tan\theta = 5sec\theta$

On squaring both sides, we get

$\rm :\longmapsto\:(4 + 3tan\theta) {}^{2} = (5sec\theta) {}^{2}$

$\rm :\longmapsto\: {4}^{2} + {(3tan\theta)}^{2} + 2 \times 4 \times 3tan\theta = 25 {sec}^{2}\theta$

$\rm :\longmapsto\:16 + 9 {tan}^{2}\theta + 24tan\theta = 25(1 + {tan}^{2}\theta)$

$\red{\bigg \{ \because \: {sec}^{2}x - {tan}^{2}x = 1 \bigg \}}$

$\rm :\longmapsto\:16 + 9 {tan}^{2}\theta + 24tan\theta = 25 + 25{tan}^{2}\theta$

$\rm :\longmapsto\:16{tan}^{2}\theta - 24tan\theta + 9 = 0$

$\rm :\longmapsto\:16{tan}^{2}\theta - 12tan\theta - 12tan\theta+ 9 = 0$

$\rm :\longmapsto\:4tan\theta(4tan\theta - 3) - 3(4tan\theta - 3) = 0$

$\rm :\longmapsto\:(4tan\theta - 3)(4tan\theta - 3) = 0$

$\bf\implies \:tan\theta = \dfrac{3}{4}$

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 Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1