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.