Question & Answer

Show that $\tan {{35}^{\circ }}\tan {{60}^{\circ }}\tan {{55}^{\circ }}\tan {{30}^{\circ }}=1$

ANSWER Verified Verified
Hint: Put the values of $\tan {{30}^{\circ }},tan{{60}^{\circ }}$ as $\dfrac{1}{\sqrt{3}},\sqrt{3}$ respectively to the expression in LHS use the identity $\tan \left( 90-\theta \right)=\cot \theta $ for converting any $\tan $ function to $\cot $ function. And use the relation between $\tan ,\cot $ functions to solve it further, which is given as $\tan \theta .\cot \theta =1$

Complete step-by-step answer:
So, we have to prove the expression;
$\tan {{35}^{\circ }}\tan {{60}^{\circ }}\tan {{55}^{\circ }}\tan {{30}^{\circ }}=1............\left( i \right)$
Here, we can observe that values of $\tan {{30}^{\circ }},\tan {{60}^{\circ }}$ can be put directly to the given expression of the left hand side. So, values of $\tan {{30}^{\circ }},\tan {{60}^{\circ }}$ is given as
$\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}},\tan {{60}^{\circ }}=\sqrt{3}...............\left( ii \right)$
Hence, LHS of equation (i) can be given as
$LHS=\tan {{35}^{\circ }}\tan {{60}^{\circ }}\tan {{55}^{\circ }}\tan {{30}^{\circ }}$
Put values of $\tan {{60}^{\circ }},\tan {{30}^{\circ }}$ from the equation (ii) to the above expression. So, we get
  & LHS=\tan {{35}^{\circ }}\times \sqrt{3}\times \tan {{55}^{\circ }}\times \dfrac{1}{\sqrt{3}} \\
 & LHS=\tan {{35}^{\circ }}\times \tan {{55}^{\circ }}\times \dfrac{\sqrt{3}}{\sqrt{3}} \\
 & LHS=\tan {{35}^{\circ }}\tan {{55}^{\circ }}...............\left( iii \right) \\
As we do not know the exact values of $\tan {{35}^{\circ }},\tan {{55}^{\circ }}$ so, we need to apply some identity by observing the equation (iii). So, we can observe that sum of angles ${{35}^{\circ }},{{55}^{\circ }}$ is ${{90}^{\circ }}\left( {{35}^{\circ }}+{{55}^{\circ }}={{90}^{\circ }} \right)$ Hence, we can convert one of the $\tan $ function to $\cot $ by applying the relation $\tan \left( 90-\theta \right)=\cot \theta $ So, we have
$\tan \left( 90-\theta \right)=\cot \theta ..................\left( iv \right)$
Put $\theta ={{35}^{\circ }}$ to the above equation so, we get
  & \tan \left( 90-35 \right)=\cot {{35}^{\circ }} \\
 & \tan 55=\cot {{35}^{\circ }}.................\left( v \right) \\
Now, out the value of $\tan {{55}^{\circ }}$ from the equation (v) to the expression (iii). Hence, we get
$LHS=\tan {{35}^{\circ }}\cot {{35}^{\circ }}.............(vi)$
Now, we know the relation between $\tan \theta ,\cot \theta $ can be given as
  & \tan \theta =\dfrac{1}{\cot \theta } \\
 & \tan \theta .\cot \theta =1 \\
So, we can get
  & LHS=\dfrac{1}{\cot {{35}^{\circ }}}\times \cot {{35}^{\circ }} \\
 & LHS=1 \\
Hence, we get LHS = RHS =1. So, the given expression is proved.

Note: We need not to put the value of \[\tan {{30}^{\circ }},\tan {{60}^{\circ }}\] in the given expression. The sum of them is \[{{90}^{\circ }}\] . So, we can use the identity $\tan \left( 90-\theta \right)=\cot \theta ,\tan \theta .\cot \theta =1$ to get \[\tan {{30}^{\circ }}.\tan {{60}^{\circ }}=1\] .
One may go wrong if he or she tries to find out the values of \[\tan {{35}^{\circ }},\tan {{55}^{\circ }}\]. As we don’t need values of them to simplify the relation. Hence just try to observe only the relation among the angles and it is the key point of the question as well.
Identity $\tan \left( 90-\theta \right)=\cot \theta $ is proved with the help of the quadrants and sign of these functions in each quadrant as well. So, be clear with the identities of these types where we subtract or add any multiple of $\pi ,\dfrac{\pi }{2}\left( {{180}^{\circ }},{{90}^{\circ }} \right)$ with the angles of the trigonometric functions. Hence, be clear with these identities to solve these problems.