How do you simplify the expression \[\sin t.\tan t - \cos ect.\tan t\]?

Hint: In order to proceed with the problem above we will use trigonometric functions as well as identities. First we will convert the tan function into the ratio of sin and cos function. Then the denominator will have cos function in common. Then using the standard identity of trigonometry, we will find the correct answer.

Complete step by step solution:
Given that,
\[\sin t.\tan t - \cos ect.\tan t\]
Now \[\tan t\] can also be written as \[\dfrac{{\sin t}}{{\cos t}}\]
\[ \Rightarrow\sin t.\dfrac{{\sin t}}{{\cos t}} - \cos ect.\dfrac{{\sin t}}{{\cos t}}\]
Also we can write the cosec function as reciprocal of sin function
\[ = \sin t.\dfrac{{\sin t}}{{\cos t}} - \dfrac{1}{{\sin t}}.\dfrac{{\sin t}}{{\cos t}}\]
Now cancelling the sin function we get,
\[\Rightarrow\sin t.\dfrac{{\sin t}}{{\cos t}} - \dfrac{1}{{\cos t}}\]
\[\Rightarrow\dfrac{{{{\sin }^2}t}}{{\cos t}} - \dfrac{1}{{\cos t}}\]
Since the denominator is same, we can write that,
\[ \Rightarrow \dfrac{{{{\sin }^2}t - 1}}{{\cos t}}\]
We know that \[{\sin ^2}t + {\cos ^2}t = 1\]
Thus we can write above expression as,
\[{\sin ^2}t - 1 = - {\cos ^2}t\]
Now the expression above will be,
\[ = \dfrac{{ - {{\cos }^2}t}}{{\cos t}}\]
The numerator can be written as,
\[ = \dfrac{{ - \cos t.\cos t}}{{\cos t}}\]
On cancelling the same terms from the numerator and denominator we get,
\[ = - \cos t\]
Thus the correct answer is
\[\sin t.\tan t - cosect.\tan t = - \cos t\]

Note: Here note that we can observe the tan function common in both the terms. That also can be taken common and proceed the same way as we did above. The trigonometric identities can be rearranged in order to reach the answer. Like in the above case we used the trigonometric identity \[{\sin ^2}t + {\cos ^2}t = 1\] .