Question

If $\dfrac{{ - \pi }}{2} < \theta < \dfrac{\pi }{2}$ and $\theta = \pm \dfrac{\pi }{4}$ then the value of $\cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right)$ is
A) 0
B) -1
C) 1
D) 2
E) -2

Answer 1

Hint : Here the question is related to the trigonometric topic. The given term is in the form of the product of the two trigonometric functions. As we know that the formula $\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A - \cot B}}$, by using this we are going to simplify the given question and we choose the appropriate the option.

Complete step-by-step answer :
The question is related to trigonometry and it includes the trigonometry ratios. The trigonometry ratios are sine, cosine, tangent, cosecant, secant, and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot.
Now consider the given question.
$\Rightarrow \cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right)$
The two trigonometric ratios are in the form of a product.
As we know, the formula $\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A - \cot B}}$. On using this formula the given term is written as
$\Rightarrow \dfrac{{\cot \dfrac{\pi }{4}\cot \theta - 1}}{{\cot \dfrac{\pi }{4} + \cot \theta }} \times \dfrac{{\cot \dfrac{\pi }{4}\cot \theta + 1}}{{\cot \dfrac{\pi }{4} - \cot \theta }}$
On considering the table of trigonometry ratio of cotangent for standard angles.

Angle	$0$	$\dfrac{\pi }{6}$	$\dfrac{\pi }{4}$	$\dfrac{\pi }{3}$	$\dfrac{\pi }{2}$
cotangent	$\infty$	$\sqrt 3$	$1$	$\dfrac{1}{{\sqrt 3 }}$	0

Now the given term is written as
$\Rightarrow \dfrac{{1.\cot \theta - 1}}{{1 + \cot \theta }} \times \dfrac{{1.\cot \theta + 1}}{{1 - \cot \theta }}$
On simplifying we have
$\Rightarrow \dfrac{{\cot \theta - 1}}{{1 + \cot \theta }} \times \dfrac{{\cot \theta + 1}}{{1 - \cot \theta }}$
In the first term, take a minus sign as a common in the numerator
$\Rightarrow \dfrac{{ - (1 - \cot \theta )}}{{1 + \cot \theta }} \times \dfrac{{\cot \theta + 1}}{{1 - \cot \theta }}$
On cancelling the terms, we have
$\Rightarrow - 1$
Therefore the value of $\cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right)$ is -1.
Hence option (B) is the correct one.
So, the correct answer is “Option B”.

Note : In trigonometry to find the value of angles we have a table of trigonometry ratios for the standard angles. The ASTC rule is applicable for the highest values. Whether the value of angle is in degree or radians the value for the standard angles will not change. Where ASTC rule is abbreviated as ALL SINE TANGENT COSINE.