Question

If tan x + 1 = √2, show that cos x – sin x = √2 sin ​

Answer 1

$\bold{\pink{\underline{Given}}}\longrightarrow \\ tanx + 1 = \sqrt{2}$

$\bold{\blue{\underline{To \: prove}}}\longrightarrow \\ cosx - sinx = \sqrt{2} sinx$

$\bold{\green{\underline{Concept \: used}}} \\ 1)tan \alpha = \dfrac{sin \alpha }{cos \alpha } \\ 2) {x}^{2} - {y}^{2} = (x + y)(x - y)$

$\bold{\red{\underline{Proof}}}\longrightarrow \\ tanx + 1 = \sqrt{2} \\ = > \dfrac{sinx}{cosx} + 1 = \sqrt{2} \\ = > \dfrac{sinx + cosx}{cosx} = \sqrt{2} \\ = > sinx + cosx = \sqrt{2} cosx \\ = > sinx = \sqrt{2} cosx - cosx\\ = > sinx = ( \sqrt{2} - 1)cosx \\ multiplying \: both \: sides \: by \: ( \sqrt{2} + 1) \\ = > ( \sqrt{2} + 1)sinx = ( \sqrt{2} + 1)( \sqrt{2} - 1)cosx \\ = > \sqrt{2} sinx + sinx = ( {( \sqrt{2}) }^{2} - {(1)}^{2} )cosx \\ = > \sqrt{2} sinx + sinx = (2 - 1)cosx \\ = > \sqrt{2} sinx = cosx - sinx \\ = > cosx - sinx = \sqrt{2} sinx$

$\bold{\pink{\underline{Additional \: iformation}}}\longrightarrow \\ 1) {sin}^{2} x + {cos}^{2} x = 1 \\ 2)1 + {tan}^{2} x = {sec}^{2} x \\ 3)1 + {cot}^{2} x = {cosec}^{2} x$

Answer 2

Given:

$\tan(x) + 1 = \sqrt{2}$

To Prove:

$\cos(x) - \sin(x) = \sqrt{2} \sin(x)$

Proof:

$\tan(x) + 1 = \sqrt{2} \\ \\ \implies \frac{ \sin(x) }{ \cos(x) } + 1 = \sqrt{2} \\ \\ \implies \: \frac{ \sin(x) + \cos(x) }{ \cos(x) } = \sqrt{2 } \\ \\ \implies \: \sin(x) + \cos(x) = \sqrt{2} \cos(x) \\ \\ \implies \: \: \sin(x) = \sqrt{2} \cos(x) - \cos(x) \\ \\ \implies \: \sin(x) = ( \sqrt{2} - 1) \cos(x ) \\ \\ \implies \: ( \sqrt{2} + 1) \sin(x) = ( \sqrt{2 } + 1)( \sqrt{2} - 1) \cos(x) \\ \\ \implies \: \sqrt{2} \sin(x) + \sin(x) = (2 - 1) \cos(x) \\ \\ \implies \: \sqrt{2} \sin(x) = \cos(x) - \sin(x) \\ \\ \implies \: \: \cos(x) - \sin(x) = \sqrt{2} \sin(x)$