Question

How do you find the value of $\cot 135$?

Answer 1

Hint: Start by converting the $\cot $ term to the simpler terms which involve $\sin ,\cos $. Then apply the formulas to simplify the equation. then next substitute the values of the terms and then further evaluate the value of $\cot 135$.

Complete step by step answer:
First we will start off by converting $\cot $ to simpler form that is the form involving $\sin ,\cos $.
We know that $\cot x = \dfrac{{\cos x}}{{\sin x}}$. Hence, now we convert $\cot $.
$\cot 135 = \dfrac{{\cos 135}}{{\sin 135}}$
Now we know that $\cos (A + B) = \cos AcosB - \sin A\sin B$ and $\sin (A + B) = \sin AcosB + \cos A\sin B$
Also, we can also write $\cos 135 = \cos (90 + 45)$ .
Now, here we will use the above mentioned formula. Hence, the expression will become,
$
   = \cos (135) \\
   = \cos (90 + 45) \\
   = \cos 90cos45 - \sin 90\sin 45 \\
   = (0)\left( {\dfrac{1}{{\sqrt 2 }}} \right) - (1)\left( {\dfrac{1}{{\sqrt 2 }}} \right) \\
   = - \left( {\dfrac{1}{{\sqrt 2 }}} \right) \\
 $
Now we calculate the value of $\sin 135$.
$
   = \sin (135) \\
   = \sin (90 + 45) \\
   = \sin 90cos45 + \cos 90\sin 45 \\
   = (1)\left( {\dfrac{1}{{\sqrt 2 }}} \right) + (0)\left( {\dfrac{1}{{\sqrt 2 }}} \right) \\
   = \left( {\dfrac{1}{{\sqrt 2 }}} \right) \\
 $
Now substitute the values of the terms in the equation $\cot 135 = \dfrac{{\cos 135}}{{\sin 135}}$.
$\cot 135 = \dfrac{{\cos 135}}{{\sin 135}} = \dfrac{{ - \dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}} = - 1$

Hence, the value of the term $\cot 135$ is $ - 1$.

Additional Information:
Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant and cosecant functions. They are also termed as arcus functions, anti-trigonometric functions or cyclometric functions.

Note: While choosing the side to solve, always choose the side where you can directly apply the trigonometric identities. Also, remember the formulas $\cos (A + B) = \cos AcosB - \sin A\sin B$ and $\sin (A + B) = \sin AcosB + \cos A\sin B$. While opening the brackets make sure you are opening the brackets properly with their respective signs. Also remember that $\cot x = \,\dfrac{{\cos x}}{{\sin x}}$.
While applying the double angle identities, first choose the identity according to the terms you have then choose the terms from the expression involving which you are using the double angle identities. While modifying any identity make sure that when you back trace the identity, you get the same original identity.