Question

Choose the correct option. Justify your choice.
$\left( {1 + \tan \theta + \sec \theta } \right)\left( {1 + \cot \theta - \cos ec\theta } \right) = $
A) $0$
B) $1$
C) $2$
D)$ - 1$

Answer 1

Hint:In order to solve the question, we have to write all the terms such as $\tan \theta $ , $\sec \theta $ ,$\cot \theta $ and $\cos ec\theta $ in terms of $\sin \theta $ and $\cos \theta $ .Use the algebraic identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ and trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$, simplify the equation and get the answer.

Complete step-by-step answer:
Let consider $T = \left( {1 + \tan \theta + \sec \theta } \right)\left( {1 + \cot \theta - \cos ec\theta } \right)$
Now we know that,
$
  \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}{\text{ , sec}}\theta {\text{ = }}\dfrac{1}{{\cos \theta }} \\
  \cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}{\text{ , }}\cos ec\theta = \dfrac{1}{{\sin \theta }} \\
 $
Using all the conversions we get,
$T = \left( {1 + \dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{1}{{\cos \theta }}} \right)\left( {1 + \dfrac{{\cos \theta }}{{\sin \theta }} - \dfrac{1}{{\sin \theta }}} \right)$
Now, we need to simplify this equation
$
   \Rightarrow T = \left( {\dfrac{{\cos \theta + \sin \theta + 1}}{{\cos \theta }}} \right)\left( {\dfrac{{\sin \theta + \cos \theta - 1}}{{\sin \theta }}} \right) \\
   \Rightarrow T = \dfrac{{\left( {\sin \theta + \cos \theta + 1} \right)\left( {\sin \theta + \cos \theta - 1} \right)}}{{\sin \theta \cos \theta }} \\
 $
Now if you carefully observe, you may see that we can use the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ where $a = \sin \theta + \cos \theta $ and $b = 1$
$ \Rightarrow T = \dfrac{{{{\left( {\sin \theta + \cos \theta } \right)}^2} - {{\left( 1 \right)}^2}}}{{\sin \theta \cos \theta }}$
Using the identity ${\left( {a + b} \right)^2} = {a^2} + {b^2}+2ab$ , where $a = \sin \theta $ and $b = \cos \theta $ , we get
$T = \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta + 2\sin \theta \cos \theta - 1}}{{\sin \theta \cos \theta }}$
Using the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ , we get
$
  T = \dfrac{{1 + 2\sin \theta \cos \theta - 1}}{{\sin \theta \cos \theta }} \\
   \Rightarrow T = \dfrac{{2\sin \theta \cos \theta }}{{\sin \theta \cos \theta }} = 2 \\
   \Rightarrow T = 2 \\
$
So, the correct answer is “Option C”.

Note:The first approach that comes to our mind by watching the equation is to open the brackets and multiply the terms. This is a very good approach but will create a lot of terms and increase the chances of mistakes. So, to avoid that we will convert every term into $\sin \theta $ and $\cos \theta $ at first.This question can also be solved by using the triangular trigonometric identities such as $\tan \theta = \dfrac{P}{B}{\text{ , cot}}\theta {\text{ = }}\dfrac{B}{P}{\text{ , }}\sec \theta = \dfrac{H}{B}{\text{ , }}\cos ec\theta = \dfrac{H}{P}$ , where $P$ is the perpendicular, $B$ is the base and $H$ is the hypotenuse. Another property is also used ${P^2} + {B^2} = {H^2}$ for this method.