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If cos θ + sin θ =√2 cosθ, show that cosθ –sinθ = √2sinθ |
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Answer» GIVEN: \cos\theta+\sin\theta=\sqrt{2}\cos\theta Squaring on both the SIDES, we get (\cos\theta+\sin\theta)^2=(\sqrt{2}\cos\theta)^2\\\\\Rightarrow\ \cos^2\theta+\sin^2\theta+2\cos\theta\sin\theta=2\cos^2\theta\\\\\Rightarrow\ 1+2\cos\theta\sin\theta=2\cos^2\theta\\\\\Rightarrow\ 2\cos\theta\sin\theta=2\cos^2\theta-1 Now, consider (\cos\theta-\sin\theta)^2=\cos^2\theta+\sin^2\theta-2\cos\theta\sin\theta\\\\=1-(2\cos^2\theta-1)\\\\=2-2cos^2\theta=2-2(1-\sin^2\theta)\\\\=2-2+2\sin^2\theta\\\\=2\sin^2\theta\\\\\Rightarrow\ (\cos\theta-\sin\theta)^2=2\sin^2\theta\\\\\Rightarrow\ \cos\theta-\sin\theta=\sqrt{2}\sin\theta Hence, proved. follow me please... MARK as BRAINLIEST please... |
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