Question

if sinA+cosA=√3 then prove that tanA + cotA=1 ​

Answer 1

Solution :

Given :

sinA + cosA = √3

Sqaring on both sides

⇒ (sinA + cosA)² = (√3)²

⇒ sin²A + cos²A + 2sinA.cosA = 3

[ Because (a + b)² = a² + b² + 2ab ]

⇒ 1 + 2sinA.cosA = 3

[ Because sin²A + cos²A = 1 ]

⇒ 2sinA.cosA = 3 - 1

⇒ 2sinA.cosA = 2

⇒ sinA.cosA = 2/2 = 1

Now, tanA + cotA

$\rm = \dfrac{sinA}{cosA} + \dfrac{cosA}{sinA}$

[ Because tanA = sinA/cosA and cotA = cosA/sinA ]

Taking LCM

$\rm = \dfrac{sinA( sinA) +cos A(cosA)}{sinA.cosA}$

$\rm = \dfrac{sin^{2} A+cos^{2} A}{sinA.cosA}$

$\rm = \dfrac{1}{sinA.cosA}$

[ Because sin²A + cos²A = 1 ]

$\rm = \dfrac{1}{1}$

[ Proved sinA.cosA = 1]

$\rm = 1$

i.e tanA + cotA = 1

Hence proved.

Answer 2

Given :----

• SinA + cosA = √3

Prove :-----

• tanA + cotA = 1

Formula used :----

• (a+b)² = a² + b² + 2ab
• sin²A + cos²A = 1
• tan A = sinA/cosA
• cot A = cosA /sinA

Solution :------

Sin A + cos A = √3

squaring both sides we get, ,

(sinA + cosA)² = (√3)²

using (a+b)² = a² + b² + 2ab now, in LHS,

sin²A + cos²A + 2sincosA = 3

using sin²A + cos²A = 1 now , we get,

→ 1 + 2 sinAcosA = 3

taking 1 , RHS we get,

→ 2 sinAcosA = 3-1

→ sinAcosA = 2/2 = 1 ------------------------------- Equation(1)

Now, solving tanA + cotA = 1 by putting value of them in terms of sinA , cos A we get,

$\frac{ \sin(a) }{ \cos(a) } + \frac{ \cos(a) }{ \sin(a) } = 1 \\ \\ taking \: lcm \: of \: denominator \\ \\ \\ \frac{ \sin^{2} (a) + \cos^{2} (a) }{\cos(a)\sin(a)} = 1 \\ \\ putting \: value \: of \: equation \: 1 \: in \: denominator \\ and \: formula \: in \: numerator \: we \: get \\ \\ \frac{1}{1} = 1 \\ \\ 1 = 1$

Hence , proved .....