The value of $$\tan {{1}^{\circ }}.\tan {{2}^{\circ }}.\tan {{3}^{\circ }}.\tan {{4}^{\circ }}..........\tan {{89}^{\circ }}$$ is:A. 0B. 1C. 2D. $$\dfrac{1}{2}$$

Question

Vedantu Content Team · Accepted Answer

Hint: First we have to confirm the value of $$\tan \left( {{90}^{\circ }}-\theta  \right)=\cot \theta $$. And then use this value on the problem by replacing $$\tan {{87}^{\circ }}.\tan {{88}^{\circ }}.\tan {{89}^{\circ }}$$ with $$\tan {{\left( 90-3 \right)}^{\circ }}.\tan {{\left( 90-2 \right)}^{\circ }}.\tan {{\left( 90-1 \right)}^{\circ }}$$. Then we have to use the fundamental multiplication formulae $$\tan \theta .\cot \theta =1$$. And in the middle position there is a $$\tan {{45}^{\circ }}$$ hidden. We have to put the value of $$\tan {{45}^{\circ }}=1$$ and calculate the rest of the equation accordingly.Complete step by step solution:In trigonometric ratios of angles $$\left( {{90}^{\circ }}-\theta  \right)$$ we can find the relation between all six trigonometric ratios.Let a rotating line OA rotate about O in the anti-clockwise direction, from initial position to ending position makes an angle $$\angle XOA=\theta $$. Now a point C is taken on OA and drawn CD perpendicular to OX.Again another rotating line OB rotates about O in the anti-clockwise direction, from initial position to ending position (OX) makes an angle $$\angle XOY={{90}^{\circ }}$$. This rotating line now rotates in the clockwise direction, starting from the position (OY) makes an angle $$\angle YOB=\theta $$.Now, we can observe that $$\angle XOB=\left( 90-\theta  \right)$$.Again a point E is taken on OB such that OC = OE and draws EF perpendicular to OX.Since, $$\angle YOB=\angle XOA$$. Therefore $$\angle OEF=\angle COD$$.Now, from the right-angled $$\vartriangle EOF$$ and right-angled $$\vartriangle COD$$ we get, $$\angle OEF=\angle COD$$ and $$OE=OC$$.Hence $$\vartriangle EOF\cong \vartriangle COD$$ (congruent).\n    \n    \n    \n    \n  In this diagram FE and OD both are positive. Similarly, OF and DC are both positive.According to the definition of trigonometric ratio we get,$$\tan \left( {{90}^{\circ }}-\theta  \right)=\cot \theta $$Now we can write the question as$$\tan {{1}^{\circ }}.\tan {{2}^{\circ }}.\tan {{3}^{\circ }}......\tan {{87}^{\circ }}.\tan {{88}^{\circ }}.\tan {{89}^{\circ }}$$In the middle position there is a $$\tan {{45}^{\circ }}$$is hidden,$$\begin{align}  & \Rightarrow \tan {{1}^{\circ }}.{{\tan }^{\circ }}\tan {{3}^{\circ }}...\tan {{45}^{\circ }}..\tan {{\left( 90-3 \right)}^{\circ }}.\tan {{\left( 90-2 \right)}^{\circ }}.\tan {{\left( 90-1 \right)}^{\circ }} \\  & \Rightarrow \tan {{1}^{\circ }}.\tan {{2}^{\circ }}.\tan {{3}^{\circ }}.....\tan {{45}^{\circ }}....\cot {{3}^{\circ }}.\cot {{2}^{\circ }}.\cot {{1}^{\circ }} \\  & \Rightarrow \tan {{1}^{\circ }}.\cot {{1}^{\circ }}.\tan {{2}^{\circ }}.\cot {{2}^{\circ }}.\tan {{3}^{\circ }}.\cot {{3}^{\circ }}....\tan {{44}^{\circ }}.\cot {{44}^{\circ }}.\tan {{45}^{\circ }} \\ \end{align}$$We know the value of $$\tan {{45}^{\circ }}=1$$ and $$\tan \theta .\cot \theta =1$$ we can write$$\begin{align}  & \Rightarrow 1\times 1\times 1\times ....\times 1 \\  & \Rightarrow 1 \\ \end{align}$$The value of $$\tan {{1}^{\circ }}.\tan {{2}^{\circ }}.\tan {{3}^{\circ }}.\tan {{4}^{\circ }}..........\tan {{89}^{\circ }}$$ is 1(option B).Note: Students have to remember the value of $$\tan {{45}^{\circ }}=1$$  and $$\tan \theta .\cot \theta =1$$. Students have to understand how to change $$\tan {{87}^{\circ }}.\tan {{88}^{\circ }}.\tan {{89}^{\circ }}$$into $$\cot {{3}^{\circ }}.\cot {{2}^{\circ }}.\cot {{1}^{\circ }}$$. This step is key to solve this type of question. They have to remember trigonometric ratios of angles $$\left( {{90}^{\circ }}-\theta  \right)$$ and its relations with trigonometric ratios.

The value of tan⁡1∘.tan⁡2∘.tan⁡3∘.tan⁡4∘..........tan⁡89∘ is:A. 0B. 1C. 2D. 12

The value of $\tan 1^{\circ} . \tan 2^{\circ} . \tan 3^{\circ} . \tan 4^{\circ} . . . . . . . . . . \tan 89^{\circ}$ is:
A. 0
B. 1
C. 2
D. $\frac{1}{2}$