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Question: Evaluate the given : \[\cos 225^\circ - \sin 225^\circ + \tan 495^\circ - \cot 495^\circ \]...

Evaluate the given : cos225sin225+tan495cot495\cos 225^\circ - \sin 225^\circ + \tan 495^\circ - \cot 495^\circ

Explanation

Solution

Here, we will use the basic identities of the trigonometric functions to find out the value of the given equation. So we will rewrite the terms given in the equation as a sum of degrees by applying the properties of the trigonometric function. Then we will use the periodicity of the trigonometric function and simplify the equation. We will then substitute the values of the function to get the value of the equation.

Complete step-by-step answer:
Let cos225sin225+tan495cot495=T\cos 225^\circ - \sin 225^\circ + \tan 495^\circ - \cot 495^\circ = T.
First, we will simplify the given equation by writing the trigonometry functions of the equation in the form of the quadrants.
We can write cos225=cos(180+45)\cos 225^\circ = \cos \left( {180^\circ + 45^\circ } \right), sin225=sin(180+45)\sin 225^\circ = \sin \left( {180^\circ + 45^\circ } \right), tan495=tan(5×90+45)\tan 495^\circ = \tan \left( {5 \times 90^\circ + 45^\circ } \right) and cot495=cot(5×90+45)\cot 495^\circ = \cot \left( {5 \times 90^\circ + 45^\circ } \right). Therefore, we get
T=cos(180+45)sin(180+45)+tan(5×90+45)cot(5×90+45)\Rightarrow T = \cos \left( {180^\circ + 45^\circ } \right) - \sin \left( {180^\circ + 45^\circ } \right) + \tan \left( {5 \times 90^\circ + 45^\circ } \right) - \cot \left( {5 \times 90^\circ + 45^\circ } \right)
Also we know that cos(180+θ)=cosθ\cos \left( {180^\circ + \theta } \right) = - \cos \theta and sin(180+θ)=sinθ\sin \left( {180^\circ + \theta } \right) = - \sin \theta as in the third quadrant both the sin and the cos function is negative.
tan(5×90+θ)=cotθ\tan \left( {5 \times 90^\circ + \theta } \right) = - \cot \theta and cot(5×90+θ)=tanθ\cot \left( {5 \times 90^\circ + \theta } \right) = - \tan \theta as in the second quadrant both the tan and the cot function is negative. Therefore the equation becomes
T=cos(45)(sin(45))cot(45)(tan(45))\Rightarrow T = - \cos \left( {45^\circ } \right) - \left( { - \sin \left( {45^\circ } \right)} \right) - \cot \left( {45^\circ } \right) - \left( { - \tan \left( {45^\circ } \right)} \right)
T=cos(45)+sin(45)cot(45)+tan(45)\Rightarrow T = - \cos \left( {45^\circ } \right) + \sin \left( {45^\circ } \right) - \cot \left( {45^\circ } \right) + \tan \left( {45^\circ } \right)
Now we will put the values of the trigonometric functions and solve this to get the value of the expression. Therefore, we get
T=12+121+1\Rightarrow T = - \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} - 1 + 1
T=12+121+1=0\Rightarrow T = - \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} - 1 + 1 = 0
cos225sin225+tan495cot495=0\Rightarrow \cos 225^\circ - \sin 225^\circ + \tan 495^\circ - \cot 495^\circ = 0
Hence, the value of the equation cos225sin225+tan495cot495\cos 225^\circ - \sin 225^\circ + \tan 495^\circ - \cot 495^\circ is 0.

Note: We should know the different properties of the trigonometric function and also in which quadrant which function is positive or negative as in the first quadrant all the functions i.e. sin, cos, tan, cot, sec, cosec is positive. In the second quadrant, only the sin and cosec function are positive and all the other functions are negative. In the third quadrant, only tan and cot function is positive and in the fourth quadrant, only cos and sec function is positive.