Question

# Evaluate the following trigonometric equation $\dfrac{\tan 65{}^\circ }{\cot 25{}^\circ }$ .

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Hint: Use conversion of trigonometric functions of cot and tan by changing their angles. Don’t go for calculating exact values of $\tan 65{}^\circ$ and $\cot 25{}^\circ$. So, convert and simplify.

Here, we need to evaluate the value of the expression $\dfrac{\tan 65{}^\circ }{\cot 25{}^\circ }$ .
As we do not know the exact value of $\tan 65{}^\circ$ and $\cot 25{}^\circ$ , so we cannot solve the expression by putting the values of $\tan 65{}^\circ$ or $\cot 25{}^\circ$. And if we go for finding values of $\tan 65{}^\circ$ and $\cot 25{}^\circ$that would be very complex and it is highly possible that we will not get exact values of them.
Hence, we need to use some relationship in $\tan \theta$ and $\cot \theta$ i.e. we need to convert $\tan \theta$ to $\cot \theta$ or vice-versa by using some identity.
As we have already learnt that we can convert one trigonometric function to another by adding $90{}^\circ$ or $180{}^\circ$ to the angle involved in the function or subtracting as well.
So, we know the complementary conversions of $\tan \theta$ to $\cot \theta$, $\sin \theta$ to $\cos \theta$, $\csc \theta$ to $\sec \theta$ or vice-versa by using identities as
$\tan \left( 90-\theta \right)=\cot \theta$ or $\cot \left( 90-\theta \right)=\tan \theta$
$\cos \left( 90-\theta \right)=\sin \theta$ or $\sin \left( 90-\theta \right)=\cos \theta$
$\sec \left( 90-\theta \right)=\csc \theta$ or $\csc \left( 90-\theta \right)=\sec \theta$
As, we have only tan and cot functions and summation of given angles i.e. 65 and 25 is $90{}^\circ$ . So, we can convert $\tan 65{}^\circ$ to cot function by the following approach.
As, we can write $\tan 65{}^\circ$ as $\tan \left( 90{}^\circ -25{}^\circ \right)$ .
Now, we can write $\tan \left( 90{}^\circ -25{}^\circ \right)$ as $\cot 25{}^\circ$ from the above identity $\tan \left( 90-\theta \right)=\cot \theta$ .
Hence, we get the given expression as $\dfrac{\tan 65{}^\circ }{\cot 25{}^\circ }=\dfrac{\cot 25{}^\circ }{\cot 25{}^\circ }=1 .$
Note: One can waste a lot of time with the trigonometric identities if he/she may go for finding exact values of $\tan 65{}^\circ$ and $\cot 25{}^\circ$. Observation of sum of both angles $65~{}^\circ +25{}^\circ =90{}^\circ$ is the key point of the question. One cannot convert $65{}^\circ$ to $180{}^\circ -115{}^\circ$ or $25{}^\circ$ to $180{}^\circ -155{}^\circ$ and now try to apply identities of $\tan \left( 180-\theta \right)$ or $\cot \left( 180-\theta \right)$ which will not give the answer. Hence writing with subtraction of $90{}^\circ$ to any of the functions is the only way to get the exact answer.