Question

How do you find the exact value of $\dfrac{{\left( {\tan {{325}^ \circ } - \tan {{25}^ \circ }} \right)}}{{\left( {1 + \tan {{325}^ \circ }\tan {{25}^ \circ }} \right)}}$ ?

Answer 1

Hint : We know that tan theta equals the side opposite theta divided by the side adjacent to theta. In other words, we can say that $\tan = \dfrac{{perpendicular}}{{base}}$ . The domain and range of tangent functions is all real numbers. Here, in this question we have to solve and find the exact value which can very easily be done by using trigonometric identities.

Complete step-by-step answer :
We all have studied various trigonometric identities. Among those, one of them is $\tan \left( {A - B} \right) = \dfrac{{\left[ {\tan \left( A \right) - \tan \left( B \right)} \right]}}{{\left[ {1 + \tan \left( A \right)\tan \left( B \right)} \right]}}$ .
Now, let us take $A = {325^ \circ }$ and $B = {25^ \circ }$ .
$\tan \left( {{{325}^ \circ } - {{25}^ \circ }} \right) = \dfrac{{\left[ {\tan {{325}^ \circ } - \tan {{25}^ \circ }} \right]}}{{\left[ {1 + \tan {{325}^ \circ }\tan {{25}^ \circ }} \right]}} \\\ \tan \left( {{{325}^ \circ } - {{25}^ \circ }} \right) = \tan {300^ \circ } \;$
We know that ${300^ \circ }$ lies in the fourth quadrant, where tangent value is negative. Also, ${300^ \circ }$ has a reference angle of ${60^ \circ }$ and we already know that $\tan {60^ \circ } = \sqrt 3$ . So, $$
$\tan {300^ \circ } = - \tan {60^ \circ } = - \sqrt 3$
Hence the exact value of $\dfrac{{\left( {\tan {{325}^ \circ } - \tan {{25}^ \circ }} \right)}}{{\left( {1 + \tan {{325}^ \circ }\tan {{25}^ \circ }} \right)}}$ is $- \sqrt 3$ .
So, the correct answer is “ $- \sqrt 3$ ”.

Note : Here, in this question we used trigonometric identity and we could easily find the value. It is highly recommended to keep all the trigonometric identities in mind while solving trigonometry questions as without using these identities, solving a question would be difficult and take a lot of time.