Question

How do you evaluate \[\tan \left( { - 20} \right) \times \cot \left( {20} \right)\] .

Answer 1

Hint: The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $\tan \left( { - \theta } \right) = - \tan \left( \theta \right)$ and $\cot \theta = \dfrac{1}{{\tan \theta }}$ . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem.

Complete step by step solution:
In the given problem, we have to simplify the product of $\tan \left( { - 20} \right)$ and \[\cot \left( {20} \right)\] .
So, \[\tan \left( { - 20} \right) \times \cot \left( {20} \right)\]
Using $\tan \left( { - \theta } \right) = - \tan \left( \theta \right)$, we get,
$ = $ \[ - \tan \left( {20} \right) \times \cot \left( {20} \right)\]
Now, we know that $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$. So, the tangent and cotangent functions are reciprocal of each other. Hence, the value of the product of tangent and cotangent is one. So, we get,
$ = $ \[ - \dfrac{{\sin \left( {20} \right)}}{{\cos \left( {20} \right)}} \times \dfrac{{\cos \left( {20} \right)}}{{\sin \left( {20} \right)}}\]
On cancelling the common terms in numerator and denominator, we get,
$ = $ \[ - 1\]
Hence, the product \[\tan \left( { - 20} \right) \times \cot \left( {20} \right)\] can be simplified as \[ - 1\] by the use of basic algebraic rules and simple trigonometric formulae.
So, the correct answer is “-1”.

Note: Trigonometric functions are also called Circular functions. Trigonometric functions are the functions that relate an angle of a right angled triangle to the ratio of two side lengths. There are $6$trigonometric functions, namely: $\sin (x)$,$\cos (x)$,$\tan (x)$,$\cos ec(x)$,$\sec (x)$and \[\cot \left( x \right)\] . Also, $\cos ec(x)$ ,$\sec (x)$and \[\cot \left( x \right)\] are the reciprocals of $\sin (x)$,$\cos (x)$and$\tan (x)$ respectively.