If $$sin\theta\times cos\theta=\frac{1}{2}$$. The value of sinθ - cosθ is where 0°< θ< 90°
Given : $$sin\theta\times cos\theta=\frac{1}{2}$$
Using, $$(sin\theta-cos\theta)^2=sin^2\theta+cos^2\theta-2(sin\theta)(cos\theta)$$
=> $$(sin\theta-cos\theta)^2=1-2(\frac{1}{2})$$
=> $$(sin\theta-cos\theta)^2=1-1=0$$
=> $$sin\theta-cos\theta=0$$
=> Ans - (A)
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