Question

Find the value of $\tan \left( 65-\theta \right)-\cot \left( 25+\theta \right)-\sec \left( 55+\theta \right)+\csc \left( 35-\theta \right)$ .

Answer 1

Hint: To find the value of the given expression, we have to convert $\tan \left( 65-\theta \right)$ into cot by using the formula $\cot \left( 90-\theta \right)=\tan \theta $ . We also have to convert $\csc \left( 35-\theta \right)$ into sec by using the formula $\csc \theta =\sec \left( 90-\theta \right)$ . Then, we have to substitute the resultant values in the given expression and simplify.

Complete step-by-step solution:
We have to find the value of $\tan \left( 65-\theta \right)-\cot \left( 25+\theta \right)-\sec \left( 55+\theta \right)+\csc \left( 35-\theta \right)$ . We know that $\cot \left( 90-\theta \right)=\tan \theta $ . Therefore, we can write $\tan \left( 65-\theta \right)$ as
$\Rightarrow \tan \left( 65-\theta \right)=\cot \left( 90-\left( 65-\theta \right) \right)=\cot \left( 25+\theta \right)...\left( i \right)$
We know that $\csc \theta =\sec \left( 90-\theta \right)$ . Therefore, we can write $\csc \left( 35-\theta \right)$ as
$\Rightarrow \csc \left( 35-\theta \right)=\sec \left( 90-\left( 35-\theta \right) \right)=\sec \left( 55+\theta \right)...\left( ii \right)$
Let us substitute (i) and (ii) in the given expression.
$\begin{align}
  & \Rightarrow \cot \left( 25+\theta \right)-\cot \left( 25+\theta \right)-\sec \left( 55+\theta \right)+\sec \left( 55+\theta \right) \\
 & =0 \\
\end{align}$
Hence, the value of $\tan \left( 65-\theta \right)-\cot \left( 25+\theta \right)-\sec \left( 55+\theta \right)+\csc \left( 35-\theta \right)$ is 0.

Note: Students must be thorough with trigonometric formulas and properties. They must note that any trigonometric function with angles $\left( 180-\theta \right),\left( 180+\theta \right),\left( 360-\theta \right)\text{ and }\left( 360+\theta \right)$ will result in the same trigonometric function with angle $\theta $ provided the sign of the function varies. All the functions will be positive at angles $\left( 90-\theta \right)$ and $\left( 360+\theta \right)$ . Sine and cosec will be positive at angles $\left( 90+\theta \right)$ and $\left( 180-\theta \right)$ . At all other angles, these will be negative. Tan and cot will be positive at angles $\left( 180+\theta \right)$ and $\left( 270-\theta \right)$ . At all other angles, these will be negative. Cos and sec will be positive at $\left( 270+\theta \right)$ and $\left( 360-\theta \right)$ . At all other angles, these will be negative.
Students can also convert $\cot \left( 25+\theta \right)$ to $\tan \left( 65-\theta \right)$ and $\sec \left( 55+\theta \right)$ to $\csc \left( 35-\theta \right)$ and simplify. This method is shown below.
We know that $\tan \left( 90-\theta \right)=\cot \theta $ . Therefore, we can write $\cot \left( 25+\theta \right)$ as
$\Rightarrow \cot \left( 25+\theta \right)=\tan \left( 90-\left( 25+\theta \right) \right)=\tan \left( 65-\theta \right)...\left( a \right)$
We know that $\csc \left( 90-\theta \right)=\sec \theta $ . Therefore, we can write $\sec \left( 55+\theta \right)$ as
$\Rightarrow \sec \left( 55+\theta \right)=\csc \left( 90-\left( 55+\theta \right) \right)=\csc \left( 35-\theta \right)...\left( b \right)$
Let us substitute (a) and (b) in the given expression.
$\begin{align}
  & \Rightarrow \tan \left( 65-\theta \right)-\tan \left( 65-\theta \right)-\csc \left( 35-\theta \right)+\csc \left( 35-\theta \right) \\
 & =0 \\
\end{align}$