Question 65

If $$sin\theta\times cos\theta=\frac{1}{2}$$. The value of sinθ - cosθ is where 0°< θ< 90°

Solution

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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