Question

What is \[\tan \,{40^ \circ }\] ?

Answer 1

Hint: Here in this question we have to find the value of \[\tan \,{40^ \circ }\], we can obtain the result directly by using the calculator. But here we write the given \[\tan \,{40^ \circ }\] in the form of sum of two angles i.e., \[\tan \,({30^ \circ } + {10^ \circ })\], then we apply the formula \[\tan \,(A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}\] and on simplification we obtain the value of \[\tan \,{40^ \circ }\].

Complete step by step answer:
The trigonometry which deals with the study of angles and the sides of the right-angled triangle. In trigonometry we have six trigonometry ratios namely, sine, cosine, tangent, cosecant, secant and cotangent. Here we must to know about the table of trigonometry ratio, it is given as,

Trigonometry ratio	\[0^ \circ }\]	\[{30^ \circ }\]	\[{45^ \circ }\]	\[{60^ \circ }\]	\[{90^ \circ }\]
Sine	0	\[\dfrac{1}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{{\sqrt 3 }}{2}\]	1
cosine	1	\[\dfrac{{\sqrt 3 }}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{1}{2}\]	0
tangent	0	\[\dfrac{1}{{\sqrt 3 }}\]	1	\[\sqrt 3 \]	\[\infty \]

Here we have to find the value of \[\tan \,{40^ \circ }\], The angle \[{40^ \circ }\] is not present in the table. So we write the angle as \[{40^ \circ } = {30^ \circ } + {10^ \circ }\].
Therefore \[\tan \,{40^ \circ }\] can be written as
\[ \Rightarrow \tan \,({30^ \circ } + {10^ \circ })\]
By the sum formula for the tangent trigonometry ratio i.e., \[\tan \,(A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}\], the above inequality is written as
\[ \Rightarrow \tan \,({30^ \circ } + {10^ \circ }) = \dfrac{{\tan {{30}^ \circ } + \tan {{10}^ \circ }}}{{1 - \tan {{30}^ \circ }.\tan {{10}^ \circ }}}\] ---- (1)
From the table of the trigonometric ratios we know the \[\tan \,{30^ \circ }\], and we write the value of \[\tan \,{10^ \circ }\] with the help of Clark’s table or by the scientific calculator.
So the value of \[\tan \,{10^ \circ } = 0.176326\].
The value of \[\tan \,{30^ \circ } = \dfrac{1}{{\sqrt 3 }} = \dfrac{1}{{1.73205}} = 0.57735\]
Substituting the known values in the equation (1) we have
\[ \Rightarrow \tan \,({30^ \circ } + {10^ \circ }) = \dfrac{{0.57735 + 0.176326}}{{1 - (0.57735).(0.176326)}}\]
On adding the terms in the numerator and multiplying the terms in the denominator we have
\[ \Rightarrow \tan \,({30^ \circ } + {10^ \circ }) = \dfrac{{0.753676}}{{1 - 0.101802}}\]
On subtracting 0.101802 from 1 we have
\[ \Rightarrow \tan \,({30^ \circ } + {10^ \circ }) = \dfrac{{0.753676}}{{0.898198}}\]
On dividing the number 0.753676 by 0.898198 we get
\[ \therefore \tan \,({40^ \circ }) = 0.839097\]
Therefore the value of \[\tan \,({40^ \circ }) = 0.839097\]

Note: The value is written directly by using the Clark’s table or the scientific calculator. In a scientific calculator first we have to set the mode in a degree mode then we have to type tan 40, then the answer will be obtained. Suppose if the calculator is in the radian mode then the obtained result will be wrong.