Question

sin A + cos A = √3, then prove that tan A + cot A = 1​

Answer 1

Answer:

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

Given:

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

To Prove:

$\tan \theta+\cot \theta=1$

Proof:

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

Squaring of both sides, we get:

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

Using the formula $(a+b)^{2}=a^{2}+b^{2}+2 a b$

Applying formula in $(\sin \theta+\cos \theta)^{2},$

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

1+2sinθcosθ=3

2sinθcosθ=2  

sinθcosθ=1  ______(1)

The value of the $\sin \theta+\cos \theta=\sqrt{3}$

is sinθcosθ=1  

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  

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

tanθ+cotθ=1=R.H.S

∴L.H.S=R.H.S

Hence proved

∴If $\bold{\sin \theta+\cos \theta=\sqrt{3}}$ then  $\bold{\tan \theta+\cot \theta=1}$ .

Answer 2

tanθ+cotθ=1 , If the value of \bold{\sin \theta+\cos \theta=\sqrt{3}}sinθ+cosθ=

3

Given:

\sin \theta+\cos \theta=\sqrt{3}sinθ+cosθ=

3

To Prove:

\tan \theta+\cot \theta=1tanθ+cotθ=1

Proof:

\sin \theta+\cos \theta=\sqrt{3}sinθ+cosθ=

3

Squaring of both sides, we get:

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

2

=(

3

)

2

Using the formula (a+b)^{2}=a^{2}+b^{2}+2 a b(a+b)

2

=a

2

+b

2

+2ab

Applying formula in (\sin \theta+\cos \theta)^{2},(sinθ+cosθ)

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}

sin

2

θ+cos

2

θ+2sinθcosθ=3

∵sin

2

θ+cos

2

θ=1

1+2sinθcosθ=3

2sinθcosθ=2

sinθcosθ=1 ______(1)

The value of the \sin \theta+\cos \theta=\sqrt{3}sinθ+cosθ=

3

is sinθcosθ=1

To prove:

tanθ+cotθ=1

L.H.S

tanθ+cotθ

Transforming the identity of tanθ ; cotθ into \frac{\sin \theta}{\cos \theta}

cosθ

sinθ

; \frac{\cos \theta}{\sin \theta}

sinθ

cosθ

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

cosθ

sinθ

+

sinθ

cosθ

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

sinθcosθ

sin

2

θ+cos

2

θ

Substituting equation (1) we get

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

1

sin

2

θ+cos

2

θ

∵sin

2

θ+cos

2

θ=1

tanθ+cotθ=1=R.H.S

∴L.H.S=R.H.S

Hence proved

∴If $\bold{\sin \theta+\cos \theta=\sqrt{3}}sin0+cosθ= 3$

$then \bold{\tan \theta+\cot \theta=1}tanθ+cotθ=1$