Question

What is the value of $$\cos \left( A+B\right) \cdot \sec \left( A-B\right) $$, If $$\cot A\cdot \cot B=2$$?
A) $$\dfrac{1}{3}$$
B) $$\dfrac{2}{3}$$
C) 1
D) -1

Answer 1

Hint: In this question it is given that if $$\cot A\cdot \cot B=2$$, then we have to find the value of $$\cos \left( A+B\right) \cdot \sec \left( A-B\right) $$. So to find the solution we first need to transform the $\sec \left( A-B\right)$ into $$\dfrac{1}{\cos \left( A-B\right) }$$ and after that we are going to use the formula-
$$\cos \left( A+B\right) =\cos A\cos B-\sin A\sin B$$........(1)
$$\cos \left( A-B\right) =\cos A\cos B+\sin A\sin B$$........(2)
Complete step-by-step solution:
Given,
$$\cos \left( A+B\right) \cdot \sec \left( A-B\right) $$
$$=\cos \left( A+B\right) \cdot \dfrac{1}{\cos \left( A-B\right) }$$[ since, $$\sec \theta =\dfrac{1}{\cos \theta }$$]
$$=\dfrac{\cos \left( A+B\right) }{\cos \left( A-B\right) }$$
$$=\dfrac{\cos A\cos B-\sin A\sin B}{\cos A\cos B+\sin A\sin B}$$[by using the formula (1) and (2)]
Now dividing the numerator and denominator by $\sin A\cdot \sin B$, we get,
$$\dfrac{\dfrac{\cos A\cos B-\sin A\sin B}{\sin A\sin B} }{\dfrac{\cos A\cos B+\sin A\sin B}{\sin A\sin B} }$$
$$=\dfrac{\dfrac{\cos A\cos B}{\sin A\sin B} -\dfrac{\sin A\sin B}{\sin A\sin B} }{\dfrac{\cos A\cos B}{\sin A\sin B} +\dfrac{\sin A\sin B}{\sin A\sin B} }$$
$$=\dfrac{\dfrac{\cos A}{\sin A} \cdot \dfrac{\cos B}{\sin B} -1}{\dfrac{\cos A}{\sin A} \cdot \dfrac{\cos B}{\sin B} +1}$$
$$=\dfrac{\cot A\cdot \cot B-1}{\cot A\cdot \cot B+1}$$ [$$\because \dfrac{\cos \theta }{\sin \theta } =\cot \theta$$]
$$=\dfrac{2-1}{2+1}$$ [ since as we know that $$\cot A\cdot \cot B=2$$]
$$=\dfrac{1}{3}$$
Hence, the correct option is option A.
Note: While simplifying a big expression, try to express it in terms of one or two basic trigonometric functions, like we have transformed the above expression in terms of $cosine$, also try to find an order in the problem to apply trigonometric identities, properties and transformations.