What is the Value of cos(50∘+A)−sin(40∘−A)cos40∘sec40∘\frac{\cos(50^\circ +A)-\sin(40^\circ -A)}{\cos40^\circ \sec40^\circ}cos40∘sec40∘cos(50∘+A)−sin(40∘−A) A0.5B1C2D0Answer: D. 0 Read Explanation: Solution:Given:cos(50∘+A)−sin(40∘−A)cos40∘sec40∘\frac{\cos(50^\circ +A)-\sin(40^\circ -A)}{\cos40^\circ \sec40^\circ}cos40∘sec40∘cos(50∘+A)−sin(40∘−A)Calculation:cos(50∘+A)−sin(40∘−A)cos40∘sec40∘\frac{\cos(50^\circ +A)-\sin(40^\circ -A)}{\cos40^\circ \sec40^\circ}cos40∘sec40∘cos(50∘+A)−sin(40∘−A)sin(90−(50∘+A))−sin(40∘−A)cos40∘sec40∘\frac{\sin(90 - (50^\circ +A))-\sin(40^\circ -A)}{\cos40^\circ \sec40^\circ}cos40∘sec40∘sin(90−(50∘+A))−sin(40∘−A)sin(40∘−A)−sin(40∘−A)cos40∘sec40∘\frac{\sin(40^\circ-A)-\sin(40^\circ -A)}{\cos40^\circ \sec40^\circ}cos40∘sec40∘sin(40∘−A)−sin(40∘−A)⇒ 0∴ The required answer is 0.$ Read more in App