Question

Find the value of \[\tan {20^ \circ } + 2\tan {50^ \circ } - \tan {70^ \circ }\] is
A) $1$
B) $0$
C) $\tan {50^ \circ }$
D) None of these

Answer 1

Hint: The trigonometric functions are real functions which relate the angle of a right angled triangle to ratios of two side lengths . There are six trigonometric ratios , they are sine , cosine , tangent , cosecant , secant and cotangent . Some formula of trigonometric function are $\tan ({90^ \circ } - \theta ) = \cot \theta $ and $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ . First we divide the largest angle by two parts and using formulas calculate that and get the required solution.

Complete step by step answer:
First we divide the $\tan {70^ \circ }$ by $\tan ({50^ \circ } + {20^ \circ })$ and we get
\[\therefore \tan {70^ \circ } = \tan ({50^ \circ } + {20^ \circ })\]
Use the formula of $\tan (A + B)$ in the above equation and we get
$ \Rightarrow \tan {70^ \circ } = \dfrac{{\tan {{50}^ \circ } + \tan {{20}^ \circ }}}{{1 - \tan {{50}^ \circ } \times \tan {{20}^ \circ }}}$
Now take cross multiplication of the above equation and we get
\[ \Rightarrow \tan {70^ \circ } - \tan {70^ \circ }\tan {50^ \circ }\tan {20^ \circ } = \tan {50^ \circ } + \tan {20^ \circ }\]
Now we write $\tan {20^ \circ } = \tan ({90^ \circ } - {70^ \circ })$ in above equation , we get
\[ \Rightarrow \tan {70^ \circ } - \tan {70^ \circ }\tan {50^ \circ }\tan ({90^ \circ } - {70^ \circ }) = \tan {50^ \circ } + \tan {20^ \circ }\]
We know the conversion $\tan ({90^ \circ } - \theta ) = \cot \theta $ , use this in the above equation and we have
$ \Rightarrow \tan {70^ \circ } - \tan {50^ \circ }\tan {70^ \circ }\cot {70^ \circ } = \tan {50^ \circ } + \tan {20^ \circ }$
Now use the formula $\tan \theta \times \cot \theta = 1$ in the above equation and we get
$ \Rightarrow \tan {70^ \circ } - \tan {50^ \circ } = \tan {50^ \circ } + \tan {20^ \circ }$
Now change the sides and signs of functions from right to left of equal sign , we have
\[ \Rightarrow \tan {70^ \circ } - \tan {50^ \circ } - \tan {50^ \circ } - \tan {20^ \circ } = 0\]
$ \Rightarrow \tan {70^ \circ } - 2\tan {50^ \circ } - \tan {20^ \circ } = 0$
Take common of $( - 1)$ sign common and we get
\[ \Rightarrow - 1 \times \left( {\tan {{20}^ \circ } + 2\tan {{50}^ \circ } - \tan {{70}^ \circ }} \right) = 0\]
Divide both sides by $( - 1)$ and get
\[ \Rightarrow \tan {20^ \circ } + 2\tan {50^ \circ } - \tan {70^ \circ } = 0\]
Therefore, the value of \[ \tan {20^ \circ } + 2\tan {50^ \circ } - \tan {70^ \circ } = 0\]. So, option (B) is correct.

Note:
Tangent is a ratio between height and base of the triangle . We know $\tan \theta = \dfrac{1}{{\cot \theta }}$ , therefore we have $\tan \theta \times \cot \theta = 1$. Try to recognize all the formulas which we used in this solution. These are very easy and useful formulas. When we divide any integers in the denominator and the numerator is zero then the answer always comes to zero.