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sinx + cosx = √2sinx Squaring the equation on both sides, we get. ⇒ sin²x + cos2 x +2sinxcosx = 2sin2 x ⇒ 1 + 2sinx.cosx = 2sin2 x ---- (i) Let sinx - cosx = k ---- (ii) Squaring the equation on both sides, we get. ⇒ 1 - 2sinxcosx = k2 Adding (i) and (ii), we get 2sin2 x + k² = 2 k² = 2(1-sin²x) k² = 2cos2 x k = ± √ (2cos2 x) The value of (sinx – cosx) = ± 2√cosx Alternate- Concept- if (sinx – cosx) =k then (sinx – cosx) =± √(2-k2) Now – k=√2 sin x k2=2sin2x (sinx – cosx) =± √(2-k2) = ± √(2-2sin2x) =± √2 (1-sin2x) = ± √2 cosx.
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