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Answer 1

Given : sinA + cosA = √2 cosA

To find : cotA

$\rule{50}2$

Solution

sinA + cosA = √2 cosA

⇒ sinA = √2 cosA - cosA

⇒ sinA = cosA(√2 - 1)

SHifting cosA to LHS,

$\sf\frac{sinA}{cosA} = \sqrt{2}-1\\\\\implies \frac{cosA}{sinA} =\frac{1}{\sqrt{2}-1}\\\\\implies cotA = \frac{1}{\sqrt{2}-1}$

(since, cosA/sinA = cotA)

Rationalizing, we get

$\sf cotA = \frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}\\\\\implies cotA = \frac{\sqrt{2}+1}{2 -1}\\\\(using, (a+b)(a-b) = a^2 - b^2)\\\\\implies cotA = \sqrt{2}+1$

Hence, cotA = √2 + 1

Answer 2

Question :-

$\star \: \small \: if \: \sin(A) + \cos(A) = \sqrt{2} \cos(A) \: \\ \\ find \: the \: value \: of \: \cot(A)$

Answer:-

cot(A) = √2 +1

Step - by - step explanation:-

Used formulas :-

$\boxed{ \star \: \cot(x) = \frac{ \cos(x) }{ \sin(x) } } \\ \\ \boxed{\star \: \tan(x) = \frac{1}{ \cot(x) } }$

Solution :-

According to the question,

$\sin(A) + \cos(A) = \sqrt{2} \cos(A) \\ \\ divided \: by \: \cos(A) \: on \: both \: sides \\ \\ \rightarrow \: \frac{ \sin(A) }{ \cos(A)} + \frac{ \cos(A) }{ \cos(A) } = \frac{ \sqrt{2} \cos(A) }{ \cos(A) } \\ \\ \rightarrow \: \: \tan(A) + 1 = \sqrt{2 } \: \\ \\ \rightarrow \: \: \tan(A) = \sqrt{2} - 1 \\ \\ \rightarrow \: \frac{1}{ \cot(A) } = \sqrt{2} - 1 \\ \\ \rightarrow \: \red{ \boxed{ \cot(A) = \frac{1} {\sqrt{2} - 1}}}$

After rationalising ,

Multiply and divided by (√2 +1)

Then we get,

cot(A) = √2 +1