Question

The value of $\sin 20^\circ \left( {\tan 10^\circ + \cot 10^\circ } \right)$.

Answer 1

This problem is based on trigonometric functions. Here the first trick which we need to apply is that we need to convert the $tan$ and $cot$ in the form of $sin$ and $cos$. Then we will take the help of some trigonometric formulas. In the end we will get our required answer.

Formula used:
We need to know the reciprocal relation $\tan x = \dfrac{1}{{\cot x}}$ and $\sec x = \dfrac{1}{{\cos x}}$.
Next, we should know the trigonometric identity ${\sec ^2}x - {\tan ^2}x = 1$.
We should also be aware of the trigonometric ratio $\tan x = \dfrac{{\sin x}}{{\cos x}}$.
The basic formula for the multiple angle identity of sine is $\sin 2x = 2\sin x\cos x$.

Complete step-by-step answer:
We have to use the different trigonometric formulas to get the required value of the given expression.
$\sin 20^\circ \left( {\tan 10^\circ + \cot 10^\circ } \right)$
We first replace $\cot 10^\circ = \dfrac{1}{{\tan 10^\circ }}$.
$= \sin 20^\circ \left( {\tan 10^\circ + \dfrac{1}{{\tan 10^\circ }}} \right)$
Now we use the formula $\sin 2x = 2\sin x\cos x$ and replace it $\sin 20^\circ = 2\sin 10^\circ \cos 10^\circ$.
$= 2\sin 10^\circ \cos 10^\circ \left( {\tan 10^\circ + \dfrac{1}{{\tan 10^\circ }}} \right)$
Next, we do L.C.M.
$= 2\sin 10^\circ \cos 10^\circ \left( {\dfrac{{{{\tan }^2}10^\circ + 1}}{{\tan 10^\circ }}} \right)$
Now we use the formula ${\sec ^2}x - {\tan ^2}x = 1$ and replace ${\tan ^2}10^\circ + 1 = {\sec ^2}10^\circ$.
$= 2\sin 10^\circ \cos 10^\circ \left( {\dfrac{{{{\sec }^2}10^\circ }}{{\tan 10^\circ }}} \right)$
Next, we use the formula $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and replace $\tan x = \dfrac{{\sin x}}{{\cos x}}$.
$= 2\sin 10^\circ \cos 10^\circ \left( {\dfrac{{{{\sec }^2}10^\circ }}{{\dfrac{{\sin 10^\circ }}{{\cos 10^\circ }}}}} \right)$
Next, we simplify the denominator and bring the terms in the numerator.
$= 2\sin 10^\circ \cos 10^\circ \times {\sec ^2}10^\circ \times \dfrac{{\cos 10^\circ }}{{\sin 10^\circ }}$
Here we use the inverse relation and replace $\sec 10^\circ = \dfrac{1}{{\cos 10^\circ }}$.
$= 2\sin 10^\circ \cos 10^\circ \times \dfrac{1}{{{{\cos }^2}10^\circ }} \times \dfrac{{\cos 10^\circ }}{{\sin 10^\circ }}$
Lastly, we perform the simplification to get the result.
$= 2 \times \dfrac{{\sin 10^\circ }}{{\sin 10^\circ }} \times \dfrac{{\cos 10^\circ \times \cos 10^\circ }}{{{{\cos }^2}10^\circ }}$
Lastly, we multiply the terms and cancel the common terms from both the numerator and denominator to get the simplified answer.
$= 2\dfrac{{\sin 10^\circ }}{{\sin 10^\circ }} \times \dfrac{{{{\cos }^2}10^\circ }}{{{{\cos }^2}10^\circ }}$
$= 2$
Hence, the required value of the expression $\sin 20^\circ \left( {\tan 10^\circ + \cot 10^\circ } \right)$ is $2$.

Note: In this type of question always remember the trigonometric properties and formulas. One should not try and substitute the given values directly in the expression to get the value for the required expression. We have to use the correct trigonometric properties and identities to simplify the terms and get a value of the expression.