Question

if tan theta +1=√2 then prove that costheta-sintheta=√2 sintheta​

Answer 1

Answer:

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Answer 2

$very \: \: easy, \: \: bro \\ use \boxed{(a - b)^{2} + (a + b)^{2} = 2(a^{2} + b^{2} )}$

Step-by-step explanation:

$Given, \: \: \: \: tan \theta + 1 = \sqrt{2} \\ Now, \: \: \: \frac{sin \theta}{cos \theta} + 1 = \sqrt{2} \\ \frac{sin \theta + cos \theta}{cos \theta} = \sqrt{2} \\ sin \theta + cos \theta = \sqrt{2} cos \theta \\ we \: \: know \: \: that \: \: \: \: \boxed{(a - b)^{2} + (a + b)^{2} = 2( {a}^{2} + {b}^{2} )} \\ putting \: \: \: a = cos \theta \: \: and \: \: b = sin \theta \\ (cos \theta - sin \theta)^{2} + (cos \theta + sin \theta)^{2} = 2(cos^{2} \theta + sin^{2} \theta )\\(cos \theta - sin \theta)^{2} + ( \sqrt{2}cos \theta )^{2} = 2 \times 1 = 2 \\ (cos \theta - sin \theta)^{2} + 2cos^{2} \theta =2 \\ (cos \theta - sin \theta)^{2} = 2 - 2cos^{2} \theta \\(cos \theta - sin \theta)^{2} = 2(1 - cos^{2} \theta ) \\(cos \theta - sin \theta)^{2} = 2sin^{2} \theta \\ cos \theta - sin \theta = \sqrt{2} sin \theta$