QUESTION

# If x = tanA – tanB, y = cotB - cotA then $\dfrac{1}{x}+\dfrac{1}{y}=$ (a)$\cot \left( A-B \right)$ (b)$\cos \left( B-A \right)$(c)$\tan \left( A-B \right)$(d)$\tan \left( B-A \right)$

Hint: First, simplify x and y individually. After simplifying, put their values in the expression given in the question and use suitable formulas to reach the answer.

Before starting with the solution to the above question, let us first discuss the important trigonometric formulas that we might use in our solution.

$\cos \left( X+Y \right)=\operatorname{cosX}\operatorname{cosY}-\operatorname{sinX}\sin Y.$

$\sin \left( X+Y \right)=\operatorname{sinX}\operatorname{cosY}+\operatorname{cosX}\sin Y.$

$\cos \left( X-Y \right)=\operatorname{cosX}\operatorname{cosY}+\operatorname{sinX}\operatorname{sinY}.$

$\sin \left( X-Y \right)=\operatorname{sinX}\operatorname{cosY}-\operatorname{cosX}\operatorname{sinY}.$

$\tan X=\dfrac{\sin X}{\cos X}$ .

$\cot X=\dfrac{\cos X}{\sin X}$ .

We will start the solution by simplifying x.

x = tanA – tanB

We know $\tan X=\dfrac{\sin X}{\cos X}$ , using this in our equation, we get

$x=\dfrac{\sin A}{\cos A}-\dfrac{\sin B}{\cos B}$

$x=\dfrac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}$

Now we will use the formula: $\sin \left( X-Y \right)=\operatorname{sinX}\operatorname{cosY}-\operatorname{cosX}\operatorname{sinY }$ , on doing so, we get

$x=\dfrac{\sin \left( A-B \right)}{\cos A\cos B}$

Similarly, now we will simplify the expression of y.

y = cotB - cotA

We know $\cot X=\dfrac{\cos X}{\sin X}$ , using this in our equation, we get

$y=\dfrac{\cos B}{\sin B}-\dfrac{\cos A}{\sin A}$

$y=\dfrac{\cos B\sin A-\sin A\cos B}{\sin A\sin B}$

Now we will use the formula: $\sin \left( X-Y \right)=\operatorname{sinX}\operatorname{cosY}-\operatorname{cosX}\operatorname{sinY }$ , on doing so, we get

$y=\dfrac{\sin \left( A-B \right)}{\sin A\sin B}$

According to the question, we need to find the value of $\dfrac{1}{x}+\dfrac{1}{y}$ .So, let us

substitute the value of x and y in the given expression using the results above, we get

$\dfrac{1}{x}+\dfrac{1}{y}$

$=\dfrac{\cos A\cos B}{\sin \left( A-B \right)}+\dfrac{\sin A\sin B}{\sin \left( A-B \right)}$

$=\dfrac{\cos A\cos B+\sin A\sin B}{\sin \left( A-B \right)}$

We know $\cos \left( X-Y \right)=\operatorname{cosX}\operatorname{cosY}+\operatorname{sinX}\operatorname{sinY }$ .On using it in our expression, we get

$=\dfrac{\cos \left( A-B \right)}{\sin \left( A-B \right)}$

We know that the above expression is equivalent to $\cot \left( A-B \right)$ . Therefore, the

Answer is option (a) $\cot \left( A-B \right)$

Note: It is preferred to keep the expression as neat as possible by eliminating all the removable terms as the larger is the number of terms in your expression more is the probability of making a mistake. In trigonometry, it is often seen that many formulas lead to the same answer. However, the complexity of the question varies with the selection of the formula. The wiser the selection of formula, the easier the question will be to solve.