Question

The value of $$\cot \left({45}^{\circ }+\theta \right).\cot \left({45}^{\circ }-\theta \right)$$ equals to
A) -1
B) 0
C) 1
D) $$\mathrm{\infty }$$

Answer 1

Hint: This problem is related to trigonometric sum and difference formula. But instead of cot function we will use tan function. Because we know the reciprocal of tangent function is cotangent. Here the formula for $$\tan \left(A+B\right)$$ & $$\tan \left(A-B\right)$$ is used. We will use the trigonometric ratio that cot is the reciprocal of tan and then further the sum and difference formula.

Complete step by step answer:
Given the expression is,
$$\cot \left({45}^{\circ }+\theta \right).\cot \left({45}^{\circ }-\theta \right)$$
Now cot can be written as reciprocal of tan function,
$$={\displaystyle \frac{1}{\tan \left({45}^{\circ }+\theta \right)}}.{\displaystyle \frac{1}{\tan \left({45}^{\circ }-\theta \right)}}$$
Now we will use the formula for $$\tan \left(A+B\right)$$ & $$\tan \left(A-B\right)$$
$$={\displaystyle \frac{1-\tan {45}^{\circ }.\tan \theta }{\tan {45}^{\circ }+\tan \theta }}.{\displaystyle \frac{1+\tan {45}^{\circ }.\tan \theta }{\tan {45}^{\circ }-\tan \theta }}$$
We know that, $$\tan {45}^{\circ }=1$$
So putting this value,
$$={\displaystyle \frac{1-\tan \theta }{1+\tan \theta }}.{\displaystyle \frac{1+\tan \theta }{1-\tan \theta }}$$
If we observe then we will come to know that the numerator of one and denominator of another have the same term. So we can cancel them,
$$=1$$
Thus, $$\cot \left({45}^{\circ }+\theta \right).\cot \left({45}^{\circ }-\theta \right)=1$$
Therefore, Option (C) is the correct answer.

Note:
Note that sin, cos and tan are the three most important or we can say pillars of the trigonometry system. Around these three only the trigonometry rotates. So we should be able to make use of the relations as well. Such that cosec, sec and cot are the reciprocals of sin, cos and tan function respectively. Also the sum and difference formula, factorization and defactorization formula are also noted for sin, cos and tan only.