Question

The value of given trigonometric expression $\cos ec 2A - \cot 2A - \tan A = $.

Answer 1

Hint- In this question, we use the concept of double angle trigonometric identities. Whenever we face a double angle in any problem so we can convert it into half of that angle. We use trigonometric identities $\sin 2A = 2\sin A\cos A{\text{ }}$ and $\cos 2A = 1 - 2{\sin ^2}A$ .

Complete step-by-step solution -
Now, we have to find value of $\cos ec 2A - \cot 2A - \tan A$
As we know we can convert double angle trigonometric terms like $\sin 2A $ and $\cos 2A $ into half of that angle like $\sin A and \cos A.$
So, first we convert $ \cos ec 2A $ and $\cot 2A$ into $\sin 2A $ and $ \cos 2A $
$
   \Rightarrow \cos ec 2A - \cot 2A - \tan A \\
   \Rightarrow \dfrac{1}{{\sin 2A}} - \frac{{\cos 2A}}{{\sin 2A}} - \tan A \\
   \Rightarrow \dfrac{{1 - \cos 2A}}{{\sin 2A}} - \tan A \\
$
We can see double angle trigonometric term so we convert into half of that angle by using identities, $\sin 2A = 2\sin A\cos A{\text{ }}$ and $\cos 2A = 1 - 2{\sin ^2}A$ .
$
   \Rightarrow \dfrac{{1 - \left( {1 - 2{{\sin }^2}A} \right)}}{{2\sin A\cos A}} - \tan A \\
   \Rightarrow \dfrac{{1 - 1 + 2{{\sin }^2}A}}{{2\sin A\cos A}} - \tan A \\
   \Rightarrow \dfrac{{2{{\sin }^2}A}}{{2\sin A\cos A}} - \tan A \\
 $
Cancel 2sinA from numerator as well as denominator,
$ \Rightarrow \dfrac{{\sin A}}{{\cos A}} - \tan A$
As we know, $\dfrac{{\sin A}}{{\cos A}} = \tan A$
$
   \Rightarrow \tan A - \tan A \\
   \Rightarrow 0 \\
$
So, the value of $\cos ec2A - \cot 2A - \tan A$ is 0.

Note- We can solve any trigonometric problem at least two ways by using different identities. First way we already mentioned above and in second way, we use trigonometric identity $\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$ and also use $\tan A = \dfrac{{\sin A}}{{\cos A}}$ .
Now, $\cos ec 2A - \cot 2A - \tan A$
We can write as, $\cos ec 2A - \left( {\dfrac{1}{{\tan 2A}} + \tan A} \right)$
We use $\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$
$
   \Rightarrow \cos ec 2A - \left( {\dfrac{{1 - {{\tan }^2}A}}{{2\tan A}} + \tan A} \right) \\
   \Rightarrow \cos ec 2A - \left( {\dfrac{{1 - {{\tan }^2}A + 2{{\tan }^2}A}}{{2\tan A}}} \right) \\
   \Rightarrow \cos ec 2A - \left( {\dfrac{{1 + {{\tan }^2}A}}{{2\tan A}}} \right) \\
$
Now use, $\tan A = \dfrac{{\sin A}}{{\cos A}}$
$ \Rightarrow \cos ec 2A - \left( {\dfrac{{{{\sin }^2}A + {{\cos }^2}A}}{{2\sin A\cos A}}} \right)$
As we know, ${\sin ^2}A + {\cos ^2}A = 1$
$ \Rightarrow \cos ec 2A - \left( {\dfrac{1}{{2\sin A\cos A}}} \right)$
We know, $\sin 2A = 2\sin A\cos A{\text{ }}$
$
   \Rightarrow \cos ec 2A - \dfrac{1}{{\sin 2A}} \\
   \Rightarrow \cos ec 2A - \cos ec2A \\
   \Rightarrow 0 \\
$