Question

Prove that: $\tan 225{}^\circ \cot 405{}^\circ +\tan 765{}^\circ \cot 675{}^\circ =0$.

Answer 1

Hint: For solving this problem, first converting the angle given in the problem statement in the range 0 to 90 degree by using suitable trigonometric properties. Now, we use the trigonometric table of values to prove the equivalence of both the sides.

Complete step-by-step answer:
Some of the important trigonometric formulas used in solving this problem are:
\[\begin{align}
& \tan \left( 360+\theta \right)=\tan \theta \\
& \tan \left( 360-\theta \right)=-\tan \theta \\
& \tan \left( 180-\theta \right)=-\tan \theta \\
& \tan \left( 180+\theta \right)=\tan \theta \\
& \cot \left( 360+\theta \right)=\cot \theta \\
& \cot \left( 360-\theta \right)=-\cot \theta \\
\end{align}\]
The specific value of functions of tan and cot which are useful are:
$\begin{align}
& \tan 45{}^\circ =1 \\
& \cot 45{}^\circ =1 \\
\end{align}$
According to the problem statement, we consider the left-hand side of the equation for proving equivalence of both sides. First, we convert the $\tan 225{}^\circ ,\cot 405{}^\circ ,\tan 765{}^\circ \text{ and }\cot 675{}^\circ $ by using the formula\[\cot \left( 360+\theta \right)=\cot \theta ,\cot \left( 360-\theta \right)=-\cot \theta ,\tan \left( 360+\theta \right)=\tan \theta \text{ and }\tan \left( 360-\theta \right)=-\tan \theta \].
$\begin{align}
& \tan 225{}^\circ =\tan \left( 360-135 \right){}^\circ \\
& \therefore \tan 225{}^\circ =-\tan \left( 135 \right){}^\circ \\
& \cot 405{}^\circ =\cot \left( 360+45 \right){}^\circ \\
& \therefore \cot 405{}^\circ =\cot \left( 45 \right){}^\circ \\
& \tan 765{}^\circ =\tan \left( 2\times 360+45 \right){}^\circ \\
& \therefore \tan 765{}^\circ =\tan \left( 45 \right){}^\circ \\
& \cot 675{}^\circ =\cot \left( 2\times 360-45 \right){}^\circ \\
& \therefore \cot 675{}^\circ =-\cot \left( 45 \right){}^\circ \\
\end{align}$
On replacing the above obtained values in the left-hand side, the required expression reduces to: $\Rightarrow -\tan 135{}^\circ \cot 45{}^\circ +\tan 45{}^\circ \left( -\cot 45{}^\circ \right)$
Now, we try to convert the above functions into the functions given in the form of table having respective angles between the range of 0 to 90 in degrees. For doing so, we use the formulas \[\tan \left( 180-\theta \right)=-\tan \theta \]. Now, we get
$\begin{align}
& \tan 135{}^\circ =\tan \left( 180-45 \right){}^\circ \\
& \therefore \tan 135{}^\circ =-\tan \left( 45 \right){}^\circ \\
\end{align}$
On replacing the above obtained values in the left-hand side, the required expression reduces to: $\begin{align}
& \Rightarrow -\left( -\tan \left( 45 \right){}^\circ \right)\cot 45{}^\circ +\tan 45{}^\circ \left( -\cot 45{}^\circ \right) \\
& \Rightarrow \tan 45{}^\circ \cot 45{}^\circ -\tan 45{}^\circ \cot 45{}^\circ \\
\end{align}$
Now, putting the values from the table, we get
$\begin{align}
& \Rightarrow 1\times 1-1\times 1 \\
& \Rightarrow 1-1 \\
& \Rightarrow 0 \\
\end{align}$
Hence, we proved the equivalence of both sides by considering the expression of the left side.

Note: Students must remember the trigonometric table and the trigonometric formulas associated with different functions. The conversion of the respective function should be done carefully, and the magnitude of the required quantity must be copied correctly in the final expression for avoiding calculation error.