Question 59

If a.sin 45°.cos 45°.tan60° = $$tan^2$$45° - cos60° then find the value of a ?

Solution

Expression : a.sin 45°.cos 45°.tan60° = $$tan^2$$45° - cos60°

=> $$a \times (\frac{1}{\sqrt{2}}) \times (\frac{1}{\sqrt{2}}) \times (\sqrt{3})=(1)^2-(\frac{1}{2})$$

=> $$\frac{\sqrt{3}}{2}a=1-\frac{1}{2}=\frac{1}{2}$$

=> $$a=\frac{1}{2} \times \frac{2}{\sqrt{3}}$$

=> $$a=\frac{1}{\sqrt{3}}$$

=> Ans - (A)

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