Question

Choose the correct option. Justify your choice
$
  \left( {\sec A + \tan A} \right)\left( {1 - \sin A} \right) = \\
  A)\,\sec A \\
  B)\,\sin A \\
  C)\,\cos ecA \\
  D)\,\cos A \\
 $

Answer 1

Hint:In order to solve the question, we have to write all the terms such as $\tan \theta $ and $\sec \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 $T = \left( {\sec A + \tan A} \right)\left( {1 - \sin A} \right)$
Now we know that $\sec A = \dfrac{1}{{\cos A}}\,\,\,\,\& \,\,\,\,\tan A = \dfrac{{\sin A}}{{\cos A}}$
Using these conversions we get
$T = \left( {\dfrac{1}{{\cos A}} + \dfrac{{\sin A}}{{\cos A}}} \right)\left( {1 - \sin A} \right)$
For easy solving we need to simplify the equation
$T = \dfrac{{\left( {1 + \sin A} \right)\left( {1 - \sin A} \right)}}{{\cos A}}$
If we observe carefully we may clearly see that we can use the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$in the numerator where $a = 1\,\,,\,\,b = \sin A$
$T = \dfrac{{1 - {{\sin }^2}A}}{{\cos A}}$
Now we see $1\,\,\& \,\,{\sin ^2}A$ in the numerator and we must check for the identity ${\sin ^2}A + {\cos ^2}A = 1$. Rearranging this we get ${\cos ^2}A = 1 - {\sin ^2}A$, using this in above equation we get
\[T = \dfrac{{{{\cos }^2}A}}{{\cos A}} = \cos A\]

So, the correct answer is “Option D”.

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 $\tan \theta = \dfrac{P}{B}{\text{ , }}\sec \theta = \dfrac{H}{B}{\text{ , }}\sin\theta = \dfrac{P}{H}$ , 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.