Question

If sine + cos o = √3

then prove that tan + cot = 1​

Answer 1

$\bold{\tan \theta+\cot \theta=1}tanθ+cotθ=1 , If \the\ value\ of\ \bold{\sin \theta+\cos \theta=\sqrt{3}}$

Given:

$\sin \theta+\cos \theta=\sqrt{3}$

To Prove:

$\tan \theta+\cot \theta$

Proof:

$\sin \theta+\cos \theta=\sqrt{3}$

Squaring of both sides, we get:

$(\sin \theta+\cos \theta)^{2}=(\sqrt{3})^{2}$

Using the formula

$(a+b)^{2}=a^{2}+b^{2}$

Applying formula in

$(\sin \theta+\cos \theta)^{2}$

$\begin{gathered}\begin{array}{l}{\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta=3} \\ {\because \sin ^{2} \theta+\cos ^{2} \theta=1}\end{array}\end{gathered}$

The value of the

$\sin \theta+\cos \theta=\sqrt{3}$

To prove:

tanθ + cotθ = 1

L.H.S

tanθ + cotθ

Transforming the identity of tanθ ; cotθ into

$\frac{\sin \theta}{\cos \theta}$

$\frac{\cos \theta}{\sin \theta}$

$\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$

$\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}$

Substituting equation (1) we get

$\begin{gathered}\begin{array}{l}{\frac{\sin ^{2} \theta+\cos ^{2} \theta}{1}} \\ {\because \sin ^{2} \theta+\cos ^{2} \theta=1}\end{array}\end{gathered}$

∴L.H.S=R.H.S

Hence proved

$∴If \bold{\sin \theta+\cos \theta=\sqrt{3}}$

$\bold{\tan \theta+\cot \theta=1}$

Thankyou :)

Answer 2

Appropriate Question :-

$\sf \: If \: sinx + cosx = \sqrt{3}, \: prove \: that \: tanx + cotx = 1$

Formula Used :-

$\purple{\boxed{ \bf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\purple{\boxed{ \bf \: {sin}^{2}x+{cos}^{2}x = 1}}$

$\purple{\boxed{ \bf \:tanx = \dfrac{sinx}{cosx}}}$

$\purple{\boxed{ \bf \:cotx = \dfrac{cosx}{sinx}}}$

$\pink{\large\underline{\bf{Solution-}}}$

Given that,

$\rm :\longmapsto\:sinx + cosx = \sqrt{3}$

On squaring both sides, we get

$\rm :\longmapsto\: {(sinx + cosx)}^{2} = {( \sqrt{3})}^{2}$

$\rm :\longmapsto\: {sin}^{2}x + {cos}^{2}x + 2sinxcosx = 3$

$\rm :\longmapsto\:1 + 2sinxcosx = 3$

$\rm :\longmapsto\:2sinxcosx = 3 - 1$

$\rm :\longmapsto\:2sinxcosx = 2$

$\rm :\longmapsto\:sinxcosx = 1$

$\red{\bf :\longmapsto\:sinxcosx = {sin}^{2}x + {cos}^{2}x}$

☆ On dividing by sinx cosx both sides, we get

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

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

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

$\rm :\implies\:\purple{\boxed{ \bf \:tanx + cotx = 1}}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

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