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Question: Show that the following relation satisfies: \(\dfrac{\cos ec\left( 90-A \right)\times \sin \left( ...

Show that the following relation satisfies:
cos⁡ec(90−A)×sin⁡(180−A)×cot⁡(360−A)sec⁡(180+A)×tan⁡(90+A)×sin⁡(−A)=1\dfrac{\cos ec\left( 90-A \right)\times \sin \left( 180-A \right)\times \cot \left( 360-A \right)}{\sec \left( 180+A \right)\times \tan \left( 90+A \right)\times \sin \left( -A \right)}=1

Explanation

Solution

Hint:To solve the above question given above, we will separately calculate the value of each term that is in multiplication and division with other terms in terms of A only. Then we will convert all the other trigonometric functions into sin and cosec functions. Then we will put the values of all the terms in the left hand side of the question and solve it.

Complete step-by-step answer:
To find the value of the left hand side of the question, we will find the value of each form that is in multiplication and division with other terms. First we will find the value of cosec (90-A). We will convert the cosec into sin function. For this, we will use the identify shown below:
cos⁡ecθ=1sin⁡θ\cos ec\theta =\dfrac{1}{\sin \theta }
Thus, we will get: cos⁡ec(90−A)=1sin⁡(90−A)\cos ec\left( 90-A \right)=\dfrac{1}{\sin \left( 90-A \right)}
Now we will use the following identity in above equation
sin⁡(90−A)=cos⁡A\sin \left( 90-A \right)=\cos A
Thus, we will get:
cos⁡ec(90−A)=1cos⁡A............(1)\cos ec\left( 90-A \right)=\dfrac{1}{\cos A}............\left( 1 \right)
Now we will convert sin⁡(180−A)\sin \left( 180-A \right) to simple form with the help of identify:
sin⁡(180−θ)=sin⁡θ\sin \left( 180-\theta \right)=\sin \theta
Thus, we will get:
sin⁡(180−A)=sinA⁡..........(2)\sin \left( 180-A \right)=\operatorname{sinA}..........\left( 2 \right)
Now we will convert the cot⁡(360−A)\cot \left( 360-A \right) into simple form with the help of following identity:
cot⁡(θ)=cot⁡(θ−360)\cot \left( \theta \right)=\cot \left( \theta -360 \right)
Thus, we will get
cot⁡(360−A)=cot⁡(360−A−360) cot⁡(360−A)=cot⁡(−A) \begin{aligned} & \cot \left( 360-A \right)=\cot \left( 360-A-360 \right) \\\ & \cot \left( 360-A \right)=\cot \left( -A \right) \\\ \end{aligned}
Now we will convert into lines and cosine terms with the help of following identity:
cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta }
Thus, we will get:
cot⁡(360−A)=cos⁡(−A)sin⁡(−A)\cot \left( 360-A \right)=\dfrac{\cos \left( -A \right)}{\sin \left( -A \right)}
Now, cosine is even functions so cos⁡(−θ)=cos⁡θ\cos \left( -\theta \right)=\cos \theta since is an odd function so  sin(−θ)=−sin⁡θ\text{ sin}\left( -\theta \right)=-\sin \theta . Thus, we will get:
 cos⁡(360−A)=cos⁡A−sin⁡A ⇒cos⁡(360−A)=−cos⁡Asin⁡A..........(3) \begin{aligned} & \ \cos \left( 360-A \right)=\dfrac{\cos A}{-\sin A} \\\ & \Rightarrow \cos \left( 360-A \right)=\dfrac{-\cos A}{\sin A}..........\left( 3 \right) \\\ \end{aligned}
Now, we will convert sec (180+A) into cosine terms with the help of following identity:
sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta }
Therefore, we get:
sec⁡(180+A)=1cos⁡(180+A)\sec \left( 180+A \right)=\dfrac{1}{\cos \left( 180+A \right)}
Now, we will convert cos⁡(180+θ)=−cos⁡θ.\cos \left( 180+\theta \right)=-\cos \theta . So we will get:
sec⁡(180+A)=1−cos⁡A..........(4)\sec \left( 180+A \right)=\dfrac{1}{-\cos A}..........\left( 4 \right)
Now, we will convert tan (90+A) into sine and cosine forms. Thus, we will get:
tan⁡(90+A)=sin⁡(90+A)cos⁡(90+A)\tan \left( 90+A \right)=\dfrac{\sin \left( 90+A \right)}{\cos \left( 90+A \right)}
Now, we know that, sin⁡(90+θ)=cos⁡θ and cos(90+θ)=−sin⁡θ\sin \left( 90+\theta \right)=\cos \theta \ \text{and cos}\left( 90+\theta \right)=-\sin \theta so we will have:
  tan⁡(90+A)=cos⁡A−sin⁡A ⇒tan⁡(90+A)=−cos⁡Asin⁡A...........(5) \begin{aligned} & \ \ \tan \left( 90+A \right)=\dfrac{\cos A}{-\sin A} \\\ & \Rightarrow \tan \left( 90+A \right)=\dfrac{-\cos A}{\sin A}...........\left( 5 \right) \\\ \end{aligned}
Now, sin⁡(−θ)\sin \left( -\theta \right) can also be written as−sin⁡θ.-\sin \theta . So, we will get:
sin⁡(−A)=−sin⁡A..........(6)\sin \left( -A \right)=-\sin A..........\left( 6 \right)
Now, we will put the values of (1),(2),(3),(4),(5)and(6)\left( 1 \right),\left( 2 \right),\left( 3 \right),\left( 4 \right),\left( 5 \right)\text{and}\left( 6 \right) into LHS of the equation. Thus, we will get:
LHS=(1cos⁡A)(sin⁡A)(−cos⁡Asin⁡A)(−1cos⁡A)(−cos⁡Asin⁡A)(−sin⁡A) ⇒LHS=(−1)(−1) ⇒LHS=1 ⇒LHS=RHS \begin{aligned} & LHS=\dfrac{\left( \dfrac{1}{\cos A} \right)\left( \sin A \right)\left( \dfrac{-\cos A}{\sin A} \right)}{\left( \dfrac{-1}{\cos A} \right)\left( \dfrac{-\cos A}{\sin A} \right)\left( -\sin A \right)} \\\ & \Rightarrow LHS=\dfrac{\left( -1 \right)}{\left( -1 \right)} \\\ & \Rightarrow LHS=1 \\\ & \Rightarrow LHS=RHS \\\ \end{aligned}
Hence proved.

Note: There is nothing given about the values of A in the question. This does not mean that we can have any value of A. At some value of A, the LHS is not defined. These values are 0,π2,π,3π2,2π........etc.0,\dfrac{\pi }{2},\pi ,\dfrac{3\pi }{2},2\pi ........\text{etc}\text{.} Thus the value of A cannot be nπ2\dfrac{n\pi }{2} because in this case LHS is not defined.