Question

If 1+tan theta=√2 then cot theta-1=

Answer 1

$\textbf{Given:}$

$1+tan\theta=\sqrt{2}$

$\implies\,tan\theta=\sqrt{2}-1$

$\text{Now,}\,cot\theta$

$=\displaystyle\frac{1}{tan\theta}$

$=\displaystyle\frac{1}{\sqrt{2}-1}{\times}\frac{\sqrt{2}+1}{\sqrt{2}+1}$

$=\displaystyle\frac{\sqrt{2}+1}{(\sqrt{2})^2-1^2}$

$=\displaystyle\frac{\sqrt{2}+1}{2-1}$

$=\sqrt{2}+1$

$\implies\,cot\theta=\sqrt{2}+1$

$\implies\,cot\theta-1=\sqrt{2}$

$\therefore\boxed{\bf\,cot\theta-1=\sqrt{2}}$

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Answer 2

Value of cotθ -1 =√2

•As, given

1 + tanθ = √2

tanθ =√2-1

• We know that

tanθ= 1/cotθ

•Now , 1/cotθ =√2-1

cotθ =1/(√2-1)

•On Rationalizing the denominator we will get

cotθ =(√2+1)/(√2-1)(√2+1)

cotθ =(√2+1)/(2-1)

cotθ =(√2+1)

Now we get , cotθ =√2+1

hence , cotθ -1 = √2