Plz tell the answer plz

1.	Plz tell the answer plz
Answer» Given Question:- $\rm \: 0 < x < 90\degree \: and \: sinx - cosx = \dfrac{1}{2}, \: then \: tanx =$ $\sf \: (1) \: \: \dfrac{1 + \sqrt{7} }{3}$ $\sf \: (2) \: \: \dfrac{8 - \sqrt{7} }{3}$ $\sf \: (3) \: \: \dfrac{4 + \sqrt{7} }{3}$ $\sf \: (4) \: \: \dfrac{\sqrt{7} - 1}{3}$ $\green{\large\underline{\sf{Solution-}}}$ Given that $\rm :\longmapsto\:sinx - cosx = \dfrac{1}{2}$ On squaring both sides, we get $\rm :\longmapsto\:(sinx - cosx)^{2} = \dfrac{1}{2}$ $\rm :\longmapsto\: {sin}^{2}x + {cos}^{2}x - 2sinxcosx = \dfrac{1}{4}$ We know, $\rm :\longmapsto\:\boxed{ \tt{ \: {sin}^{2}x + {cos}^{2}x = 1}}$ So, USING this, we get $\rm :\longmapsto\: 1 - 2sinxcosx = \dfrac{1}{4}$ $\rm :\longmapsto\: - 2sinxcosx = \dfrac{1}{4} - 1$ $\rm :\longmapsto\: - 2sinxcosx = \dfrac{1 - 4}{4}$ $\rm :\longmapsto\: - 2sinxcosx = \dfrac{- 3}{4}$ $\rm :\longmapsto\: sinxcosx = \dfrac{3}{8}$ $\rm :\longmapsto\:8sinxcosx = 3$ $\red{\rm :\longmapsto\:Divide \: both \: by \: {cos}^{2}x \: }$ $\rm :\longmapsto\:\dfrac{8sinx \: cosx}{ {cos}^{2}x} = \dfrac{3}{ {cos}^{2} x}$ $\rm :\longmapsto\:8tanx = 3 {sec}^{2}x$ We know, $\rm :\longmapsto\:\boxed{ \tt{ \: {sec}^{2}x = 1 + {tan}^{2}x \: }}$ $\rm :\longmapsto\:8tanx = 3(1 + {tan}^{2}x)$ $\rm :\longmapsto\:8tanx = 3 + 3{tan}^{2}x$ $\rm :\longmapsto\:3 {tan}^{2}x - 8tanx + 3 = 0$ So, its a quadratic EQUATION in tanx, whose solution is given by using Quadratic formula. $\boxed{ \tt{ \: tanx = \frac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a} \: }}$ Here, $\red{\rm :\longmapsto\:a = 3}$ $\red{\rm :\longmapsto\:b = - 8}$ $\red{\rm :\longmapsto\:c = 3}$ So, on substituting the values, we get $\rm :\longmapsto\:tanx = \dfrac{ - ( - 8) \: \pm \: \sqrt{ {( - 8)}^{2} - 4(3)(3)} }{2 \times 3}$ $\rm :\longmapsto\:tanx = \dfrac{8 \: \pm \: \sqrt{ 64 - 36} }{6}$ $\rm :\longmapsto\:tanx = \dfrac{8 \: \pm \: \sqrt{28} }{6}$ $\rm :\longmapsto\:tanx = \dfrac{8 \: \pm \: \sqrt{2 \times 2 \times 7} }{6}$ $\rm :\longmapsto\:tanx = \dfrac{8 \: \pm \: 2\sqrt{7} }{6}$ $\rm :\longmapsto\:tanx = \dfrac{4 \: \pm \: \sqrt{7} }{3}$ So, $\rm :\longmapsto\:tanx = \dfrac{4 \: + \: \sqrt{7} }{3} \: \: or \: \: \dfrac{4 \: - \: \sqrt{7} }{3}$ Hence, Option (3) is correct. ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬ 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

Answer»

Given Question:-

$\rm \: 0 < x < 90\degree \: and \: sinx - cosx = \dfrac{1}{2}, \: then \: tanx =$

$\sf \: (1) \: \: \dfrac{1 + \sqrt{7} }{3}$

$\sf \: (2) \: \: \dfrac{8 - \sqrt{7} }{3}$

$\sf \: (3) \: \: \dfrac{4 + \sqrt{7} }{3}$

$\sf \: (4) \: \: \dfrac{\sqrt{7} - 1}{3}$

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

Given that

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

On squaring both sides, we get

$\rm :\longmapsto\:(sinx - cosx)^{2} = \dfrac{1}{2}$

$\rm :\longmapsto\: {sin}^{2}x + {cos}^{2}x - 2sinxcosx = \dfrac{1}{4}$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: {sin}^{2}x + {cos}^{2}x = 1}}$

So, USING this, we get

$\rm :\longmapsto\: 1 - 2sinxcosx = \dfrac{1}{4}$

$\rm :\longmapsto\: - 2sinxcosx = \dfrac{1}{4} - 1$

$\rm :\longmapsto\: - 2sinxcosx = \dfrac{1 - 4}{4}$

$\rm :\longmapsto\: - 2sinxcosx = \dfrac{- 3}{4}$

$\rm :\longmapsto\: sinxcosx = \dfrac{3}{8}$

$\rm :\longmapsto\:8sinxcosx = 3$

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

$\rm :\longmapsto\:\dfrac{8sinx \: cosx}{ {cos}^{2}x} = \dfrac{3}{ {cos}^{2} x}$

$\rm :\longmapsto\:8tanx = 3 {sec}^{2}x$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: {sec}^{2}x = 1 + {tan}^{2}x \: }}$

$\rm :\longmapsto\:8tanx = 3(1 + {tan}^{2}x)$

$\rm :\longmapsto\:8tanx = 3 + 3{tan}^{2}x$

$\rm :\longmapsto\:3 {tan}^{2}x - 8tanx + 3 = 0$

So, its a quadratic EQUATION in tanx, whose solution is given by using Quadratic formula.

$\boxed{ \tt{ \: tanx = \frac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a} \: }}$

Here,

$\red{\rm :\longmapsto\:a = 3}$

$\red{\rm :\longmapsto\:b = - 8}$

$\red{\rm :\longmapsto\:c = 3}$

So, on substituting the values, we get

$\rm :\longmapsto\:tanx = \dfrac{ - ( - 8) \: \pm \: \sqrt{ {( - 8)}^{2} - 4(3)(3)} }{2 \times 3}$

$\rm :\longmapsto\:tanx = \dfrac{8 \: \pm \: \sqrt{ 64 - 36} }{6}$

$\rm :\longmapsto\:tanx = \dfrac{8 \: \pm \: \sqrt{28} }{6}$

$\rm :\longmapsto\:tanx = \dfrac{8 \: \pm \: \sqrt{2 \times 2 \times 7} }{6}$

$\rm :\longmapsto\:tanx = \dfrac{8 \: \pm \: 2\sqrt{7} }{6}$

$\rm :\longmapsto\:tanx = \dfrac{4 \: \pm \: \sqrt{7} }{3}$

So,

$\rm :\longmapsto\:tanx = \dfrac{4 \: + \: \sqrt{7} }{3} \: \: or \: \: \dfrac{4 \: - \: \sqrt{7} }{3}$

Hence, Option (3) is correct.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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

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