Question

\[3 \frac{\sin 47^\circ}{\cos 43^\circ} - \frac{\tan 50^\circ}{\cot 40^\circ} - 2\sin^2 45^\circ\] is equal to:

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Answer 1

1

Explanation:

Solving,

\[\Rightarrow 3\, \frac{\sin 47^\circ}{\cos 43^\circ} - \frac{\tan 50^\circ}{\cot 40^\circ} - 2\sin^2 45^\circ\]

\[\Rightarrow 3\, \frac{\sin 47^\circ}{\cos(90^\circ - 47^\circ)} - \frac{\tan 50^\circ}{\cot(90^\circ - 50^\circ)} - 2\sin^2 45^\circ\]

\[\Rightarrow 3\, \frac{\sin 47^\circ}{\sin 47^\circ} - \frac{\tan 50^\circ}{\tan 50^\circ} - 2 \times \left(\frac{1}{\sqrt{2}}\right)^2\]

\[\Rightarrow 3 \times 1 - 1 - 2 \times \frac{1}{2}\]

\[\Rightarrow 3 - 1 - 1\]

\[\Rightarrow 1.\]