Solveeit Logo
Login

Question

Question: Find the value of the trigonometric expression \(\dfrac{\sin \left( 18 \right)+\tan \left( 72 \right...

Find the value of the trigonometric expression sin⁡(18)+tan⁡(72)−cos⁡(72)cosec(75)+cot⁡(18)−sec⁡(15)=1\dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 72 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=1.

Explanation

Solution

Hint: We will use the trigonometric angular formula which are given by cos⁡(72)=cos⁡(90−18)\cos \left( 72 \right)=\cos \left( 90-18 \right), cos⁡(90−18)=sin⁡(18)\cos \left( 90-18 \right)=\sin \left( 18 \right), cosec(90−15)=sec⁡(15)\text{cosec}\left( 90-15 \right)=\sec \left( 15 \right) and cot⁡(18)=cot⁡(90−72)\cot \left( 18 \right)=\cot \left( 90-72 \right) to solve the given trigonometric question. By this we will be able to reduce any angle in terms of 90 degrees and after that we will cancel the common terms.

Complete step-by-step answer:
We will consider the equation sin⁡(18)+tan⁡(72)−cos⁡(72)cosec(75)+cot⁡(18)−sec⁡(15)=1\dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 72 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=1...(i).
First we will consider the left hand side of the expression we will have sin⁡(18)+tan⁡(72)−cos⁡(72)cosec(75)+cot⁡(18)−sec⁡(15)\dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 72 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}.
Now we will consider the expressioncos⁡(72)=cos⁡(90−18)\cos \left( 72 \right)=\cos \left( 90-18 \right). As we know that the value of cos⁡(90−18)=sin⁡(18)\cos \left( 90-18 \right)=\sin \left( 18 \right) therefore, we will get a new expression which is given by,
sin⁡(18)+tan⁡(72)−cos⁡(90−18)cosec(75)+cot⁡(18)−sec⁡(15)=sin⁡(18)+tan⁡(72)−sin⁡(18)cosec(75)+cot⁡(18)−sec⁡(15) ⇒sin⁡(18)+tan⁡(72)−cos⁡(90−18)cosec(75)+cot⁡(18)−sec⁡(15)=tan⁡(72)cosec(75)+cot⁡(18)−sec⁡(15) \begin{aligned} & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\sin \left( 18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)} \\\ & \Rightarrow \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)} \\\ \end{aligned}
Also if we put the value of the trigonometric term cosec(90−15)=sec⁡(15)\text{cosec}\left( 90-15 \right)=\sec \left( 15 \right) in the term of angle as 15, therefore we now have the trigonometric expression as,
sin⁡(18)+tan⁡(72)−cos⁡(90−18)cosec(75)+cot⁡(18)−sec⁡(15)=tan⁡(72)cosec(75)+cot⁡(18)−sec⁡(15) ⇒sin⁡(18)+tan⁡(72)−cos⁡(90−18)cosec(75)+cot⁡(18)−sec⁡(15)=tan⁡(72)cosec(90−15)+cot⁡(18)−sec⁡(15) sin⁡(18)+tan⁡(72)−cos⁡(90−18)cosec(75)+cot⁡(18)−sec⁡(15)=tan⁡(72)sec⁡(15)+cot⁡(18)−sec⁡(15) sin⁡(18)+tan⁡(72)−cos⁡(90−18)cosec(75)+cot⁡(18)−sec⁡(15)=tan⁡(72)cot⁡(18) \begin{aligned} & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)} \\\ & \Rightarrow \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\text{cosec}\left( 90-15 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)} \\\ & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\sec \left( 15 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)} \\\ & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\cot \left( 18 \right)} \\\ \end{aligned}
Also, as we know that the value of cot⁡(18)=cot⁡(90−72)\cot \left( 18 \right)=\cot \left( 90-72 \right) thus we will get
sin⁡(18)+tan⁡(72)−cos⁡(90−18)cosec(75)+cot⁡(18)−sec⁡(15)=tan⁡(72)cot⁡(18) sin⁡(18)+tan⁡(72)−cos⁡(90−18)cosec(75)+cot⁡(18)−sec⁡(15)=tan⁡(72)cot⁡(90−72) sin⁡(18)+tan⁡(72)−cos⁡(90−18)cosec(75)+cot⁡(18)−sec⁡(15)=tan⁡(72)tan⁡(72) sin⁡(18)+tan⁡(72)−cos⁡(90−18)cosec(75)+cot⁡(18)−sec⁡(15)=1 \begin{aligned} & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\cot \left( 18 \right)} \\\ & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\cot \left( 90-72 \right)} \\\ & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\tan \left( 72 \right)} \\\ & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=1 \\\ \end{aligned}
Hence, we have proved the desired trigonometric equation which is given by sin⁡(18)+tan⁡(72)−cos⁡(72)cosec(75)+cot⁡(18)−sec⁡(15)=1\dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 72 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=1.

Note: We can also substitute the expression sin⁡(18)=sin⁡(90−72)\sin \left( 18 \right)=\sin \left( 90-72 \right) in the expression and proceed in the same way. This will also result in the right answer. We need to take care that as we are taking the angles in terms of 90 degrees so the trigonometric terms will be the same. For example of cot⁡(18)=cot⁡(90−72)\cot \left( 18 \right)=\cot \left( 90-72 \right)instead of cot⁡(18)=tan⁡(90−72)\cot \left( 18 \right)=\tan \left( 90-72 \right). We will consider the left hand side first of the given trigonometric expression so that we will lead the left hand side to the right side of the expression. By considering these points in notice we will get the desired result.