Question

How do you find the exact value of $\tan 405$ degrees?

Answer 1

Hint : We have to find the value of $\tan {405^ \circ }$. For this, we will first write the given ${405^ \circ }$ as a sum of the known angles and then we will convert it using the trigonometric ratios of compound angles formula accordingly. Then using the values of trigonometric ratios of some standard angles we will find the result.

Complete step-by-step answer :
To solve this problem, we should know some basic terms that are:
Trigonometric Ratios of Allied angles:
Two angles are said to be allied when their sum or difference is either zero or a multiple of ${90^ \circ }$.
Trigonometric Ratios:
Trigonometric ratios are the relation between different sides and angles of a right-angled triangle.
Trigonometric Ratios of Compound angles:
Generally, the algebraic sum of two or more angles are called compound angles.
We will use the standard formula which will simplify the question i.e., $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Now, from the given question we have
$\Rightarrow \tan {405^ \circ }$
On writing in terms of compound angles, we get
$\Rightarrow \tan \left( {{{360}^ \circ } + {{45}^ \circ }} \right)$
Using the formula of $\tan \left( {A + B} \right)$ we get
$\Rightarrow \tan \left( {{{360}^ \circ } + {{45}^ \circ }} \right) = \dfrac{{\tan {{360}^ \circ } + \tan {{45}^ \circ }}}{{1 - \tan {{360}^ \circ }\tan {{45}^ \circ }}}$
As we know $\tan {360^ \circ } = 0$ and $\tan {45^ \circ } = 1$, using this we will get
$\Rightarrow \tan {405^ \circ } = \dfrac{{0 + 1}}{{1 - \left( 0 \right) \times \left( 1 \right)}}$
On solving we get
$\Rightarrow \tan {405^ \circ } = 1$
Therefore, the value of $\tan {405^ \circ }$ is $1$.
So, the correct answer is “1”.

Note : We can also solve this problem by another method.
As we know, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Therefore, using this we can write
$\Rightarrow \tan {405^ \circ } = \dfrac{{\sin {{405}^ \circ }}}{{\cos {{405}^ \circ }}}$
On writing in terms of compound angles, we get
$\Rightarrow \tan {405^ \circ } = \dfrac{{\sin \left( {{{360}^ \circ } + {{45}^ \circ }} \right)}}{{\cos \left( {{{360}^ \circ } + {{45}^ \circ }} \right)}} - - - \left( 1 \right)$
By compound angle formula we know that,
$\sin \left( {A + B} \right) = \sin A\cos B + \sin B\cos A$
$\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
Using these compound angle formulas, we can write $\left( 1 \right)$ as
$\Rightarrow \tan {405^ \circ } = \dfrac{{\sin {{360}^ \circ }\cos {{45}^ \circ } + \sin {{45}^ \circ }\cos {{360}^ \circ }}}{{\cos {{360}^ \circ }\cos {{45}^ \circ } - \sin {{360}^ \circ }\sin {{45}^ \circ }}}$
As we know, $\sin {360^ \circ } = 0$, $\cos {360^ \circ } = 1$ $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$, $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$.
Using these values, we get
$\Rightarrow \tan {405^ \circ } = \dfrac{{\left( 0 \right) \times \left( {\dfrac{1}{{\sqrt 2 }}} \right) + \left( {\dfrac{1}{{\sqrt 2 }}} \right) \times \left( 1 \right)}}{{\left( 1 \right) \times \left( {\dfrac{1}{{\sqrt 2 }}} \right) - \left( 0 \right) \times \left( {\dfrac{1}{{\sqrt 2 }}} \right)}}$
On solving we get
$\Rightarrow \tan {405^ \circ } = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}}$
On further solving we get the result as
$\Rightarrow \tan {405^ \circ } = 1$