
Find the value of \[\sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x\]
Answer
490.8k+ views
Hint:
Here, we will use the ASTC rule to simplify the sine function of the given expression. Then we will use the Trigonometric Co-ratio to simplify the given Trigonometric function and find the required value. Trigonometric identity is an equation that is always true for all the variables.
Formula Used:
We will use the following formula:
1) Trigonometric Ratio: \[\sin \left( {\pi + x} \right) = - \sin x\]
2) \[\sin \left( {\pi - x} \right) = \sin x\]
3) Trigonometric Co-ratio: \[\csc x = \dfrac{1}{{\sin x}}\]
Complete step by step solution:
We have to find the value of \[\sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x\].
We know that \[\pi + x\] lies in the third quadrant and by using the ASTC rule in Trigonometry we can say that the trigonometric ratio sine is negative in the third quadrant.
So, we get \[\sin \left( {\pi + x} \right) = - \sin x\]
Also, we know that \[\pi - x\] lies in the second quadrant and by using the ASTC rule in Trigonometry we can say that the trigonometric ratio sine is positive in the second quadrant.
So, we get \[\sin \left( {\pi - x} \right) = \sin x\]
Now we can write given equation as:
\[\sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x = - \sin x\sin x{\csc ^2}x\]
Multiplying the terms, we get
\[ \Rightarrow \sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x = - {\sin ^2}x{\csc ^2}x\]
Now, by using the Trigonometric Co-ratio \[\csc x = \dfrac{1}{{\sin x}}\], we get
\[ \Rightarrow \sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x = - {\sin ^2}x \times \dfrac{1}{{{{\sin }^2}x}}\]
Cancelling out the similar terms, we get
\[ \Rightarrow \sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x = - 1\]
Therefore, the value of \[\sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x\] is \[ - 1\].
Note:
We know that Trigonometric Equation is defined as an equation involving the trigonometric ratios. We should remember the rules that all the Trigonometric Ratios are positive in the First Quadrant. Sine and Cosecant are positive in the second quadrant and the rest are negative. Tangent and Cotangent are positive in the third quadrant and the rest are negative. Cosine and Secant are positive in the fourth quadrant and the rest are negative. This can be remembered as the ASTC rule in Trigonometry. This rule is used in determining the signs of the trigonometric ratio.
Here, we will use the ASTC rule to simplify the sine function of the given expression. Then we will use the Trigonometric Co-ratio to simplify the given Trigonometric function and find the required value. Trigonometric identity is an equation that is always true for all the variables.
Formula Used:
We will use the following formula:
1) Trigonometric Ratio: \[\sin \left( {\pi + x} \right) = - \sin x\]
2) \[\sin \left( {\pi - x} \right) = \sin x\]
3) Trigonometric Co-ratio: \[\csc x = \dfrac{1}{{\sin x}}\]
Complete step by step solution:
We have to find the value of \[\sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x\].
We know that \[\pi + x\] lies in the third quadrant and by using the ASTC rule in Trigonometry we can say that the trigonometric ratio sine is negative in the third quadrant.
So, we get \[\sin \left( {\pi + x} \right) = - \sin x\]
Also, we know that \[\pi - x\] lies in the second quadrant and by using the ASTC rule in Trigonometry we can say that the trigonometric ratio sine is positive in the second quadrant.
So, we get \[\sin \left( {\pi - x} \right) = \sin x\]
Now we can write given equation as:
\[\sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x = - \sin x\sin x{\csc ^2}x\]
Multiplying the terms, we get
\[ \Rightarrow \sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x = - {\sin ^2}x{\csc ^2}x\]
Now, by using the Trigonometric Co-ratio \[\csc x = \dfrac{1}{{\sin x}}\], we get
\[ \Rightarrow \sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x = - {\sin ^2}x \times \dfrac{1}{{{{\sin }^2}x}}\]
Cancelling out the similar terms, we get
\[ \Rightarrow \sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x = - 1\]
Therefore, the value of \[\sin \left( {\pi + x} \right)\sin \left( {\pi - x} \right){\csc ^2}x\] is \[ - 1\].
Note:
We know that Trigonometric Equation is defined as an equation involving the trigonometric ratios. We should remember the rules that all the Trigonometric Ratios are positive in the First Quadrant. Sine and Cosecant are positive in the second quadrant and the rest are negative. Tangent and Cotangent are positive in the third quadrant and the rest are negative. Cosine and Secant are positive in the fourth quadrant and the rest are negative. This can be remembered as the ASTC rule in Trigonometry. This rule is used in determining the signs of the trigonometric ratio.
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