# Find the value of $\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ $.

(a) $\dfrac{1}{16}$

(b) $0$

(c) $1$

(d) None of these

Answer

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Hint: First use the formula of $\tan \theta $ to $\cot \theta $ conversion which is: $\tan \left( 90{}^\circ -\theta \right)=\cot \theta $.

Then use the formula $tan\theta \cot \theta =1$

Complete step-by-step answer:

We need to find the value of $\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ $

We will first use the formula of $\tan \theta $ to $\cot \theta $ conversion: on either the first two terms or the last two terms.

Here, let us use it on $\tan 10{}^\circ $and $\tan 15{}^\circ $.

So $\tan 10{}^\circ $can be written as $\tan \left( 90{}^\circ -80{}^\circ \right)$

Now use the formula $\tan \left( 90{}^\circ -\theta \right)=\cot \theta $ where $\theta =80{}^\circ $

We will get the following:

$\tan \left( 90{}^\circ -80{}^\circ \right)=\cot 80{}^\circ $

So,$\tan 10{}^\circ =\cot 80{}^\circ $ …(1)

Similarly, $\tan 15{}^\circ $ can be written as $\tan \left( 90{}^\circ -75{}^\circ \right)$

Now use the formula $\tan \left( 90{}^\circ -\theta \right)=\cot \theta $ where $\theta =75{}^\circ $

We will get the following:

$\tan \left( 90{}^\circ -75{}^\circ \right)=\cot 75{}^\circ $

So, $\tan 15{}^\circ =\cot 75{}^\circ $ …(2)

Now we will substitute equations (1) and (2) in the given expression:

\[\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ =\cot 80{}^\circ \cot 75{}^\circ \tan 75{}^\circ \tan 80{}^\circ \]

Rearranging the terms, we get the following:

\[\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ =\left( \tan 80{}^\circ \cot 80{}^\circ \right)\left( \tan 75{}^\circ \cot 75{}^\circ \right)\] …(3)

Now we will use the formula $tan\theta \cot \theta =1$ which is derived by using $\left( \cot \theta =\dfrac{1}{\tan \theta } \right)$

So, \[\tan 80{}^\circ \cot 80{}^\circ =1\]

And \[\tan 75{}^\circ \cot 75{}^\circ =1\] …(4)

We will now substitute the equations in (4) to (3), we will get the following:

\[\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ =\left( \tan 80{}^\circ \cot 80{}^\circ \right)\left( \tan 75{}^\circ \cot 75{}^\circ \right)\]

\[\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ =\left( 1 \right)\left( 1 \right)\]

So, \[\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ =1\]

which is option (c)

So the final answer is (c) $1$

Note: Instead of converting the first two terms from $\tan 10{}^\circ $and $\tan 15{}^\circ $to $\cot 80{}^\circ $and $\cot 75{}^\circ $respectively, one can change the last two terms from $\tan 75{}^\circ $and $\tan 80{}^\circ $to $\cot 15{}^\circ $and $\cot 10{}^\circ $respectively too and then proceed. The final answer will be the same in both the cases.

Then use the formula $tan\theta \cot \theta =1$

Complete step-by-step answer:

We need to find the value of $\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ $

We will first use the formula of $\tan \theta $ to $\cot \theta $ conversion: on either the first two terms or the last two terms.

Here, let us use it on $\tan 10{}^\circ $and $\tan 15{}^\circ $.

So $\tan 10{}^\circ $can be written as $\tan \left( 90{}^\circ -80{}^\circ \right)$

Now use the formula $\tan \left( 90{}^\circ -\theta \right)=\cot \theta $ where $\theta =80{}^\circ $

We will get the following:

$\tan \left( 90{}^\circ -80{}^\circ \right)=\cot 80{}^\circ $

So,$\tan 10{}^\circ =\cot 80{}^\circ $ …(1)

Similarly, $\tan 15{}^\circ $ can be written as $\tan \left( 90{}^\circ -75{}^\circ \right)$

Now use the formula $\tan \left( 90{}^\circ -\theta \right)=\cot \theta $ where $\theta =75{}^\circ $

We will get the following:

$\tan \left( 90{}^\circ -75{}^\circ \right)=\cot 75{}^\circ $

So, $\tan 15{}^\circ =\cot 75{}^\circ $ …(2)

Now we will substitute equations (1) and (2) in the given expression:

\[\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ =\cot 80{}^\circ \cot 75{}^\circ \tan 75{}^\circ \tan 80{}^\circ \]

Rearranging the terms, we get the following:

\[\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ =\left( \tan 80{}^\circ \cot 80{}^\circ \right)\left( \tan 75{}^\circ \cot 75{}^\circ \right)\] …(3)

Now we will use the formula $tan\theta \cot \theta =1$ which is derived by using $\left( \cot \theta =\dfrac{1}{\tan \theta } \right)$

So, \[\tan 80{}^\circ \cot 80{}^\circ =1\]

And \[\tan 75{}^\circ \cot 75{}^\circ =1\] …(4)

We will now substitute the equations in (4) to (3), we will get the following:

\[\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ =\left( \tan 80{}^\circ \cot 80{}^\circ \right)\left( \tan 75{}^\circ \cot 75{}^\circ \right)\]

\[\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ =\left( 1 \right)\left( 1 \right)\]

So, \[\tan 10{}^\circ \tan 15{}^\circ \tan 75{}^\circ \tan 80{}^\circ =1\]

which is option (c)

So the final answer is (c) $1$

Note: Instead of converting the first two terms from $\tan 10{}^\circ $and $\tan 15{}^\circ $to $\cot 80{}^\circ $and $\cot 75{}^\circ $respectively, one can change the last two terms from $\tan 75{}^\circ $and $\tan 80{}^\circ $to $\cot 15{}^\circ $and $\cot 10{}^\circ $respectively too and then proceed. The final answer will be the same in both the cases.

Last updated date: 24th Sep 2023

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