The question asks for the value of the expression \(\frac{\sin 60^\circ}{\cos 60^\circ}\).
Recall the fundamental trigonometric identity:
\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
Applying this identity to the given expression with \(\theta = 60^\circ\):
\(\frac{\sin 60^\circ}{\cos 60^\circ} = \tan 60^\circ\)
We know the standard trigonometric values for \(60^\circ\):
Substitute these values into the expression:
\(\frac{\sin 60^\circ}{\cos 60^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\)
To simplify the division of fractions, multiply the numerator by the reciprocal of the denominator:
\(\frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3}\)
Alternatively, knowing that \(\tan 60^\circ = \sqrt{3}\) directly gives the answer.
The calculated value is \(\sqrt{3}\). This corresponds to Option A.
If \(cosec~\theta =\frac{29}{21}\) where 0 < θ < 90°, then what is the value of 4 sec θ + 4 tan θ?
The value of \(\cot \left( {cose{c^{ - 1}}\frac{5}{3} + {{\tan }^{ - 1}}\frac{2}{3}\;} \right)\)
The distance of the highest point on the graph of the function y = √3 cos x + sin x from the x-axis is:
If A = π / 6 and B = π / 3, then consider the following statements:
I. sin A + sin B = cos A + cos B
II. tan A + tan B = cot A + cot B
Which of the above statements is / are correct?
If \(\sin A = \frac{1}{{\sqrt 2 }}\) and \({\mathop{\rm Cos}\nolimits} B = \frac{{\sqrt 3 }}{2}\), then, find the value of (A + B)º.
A. 60º
B. 75º
C. 105º
D. 90º