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Question: Choose the correct option. Justify your choice \( \left( {\sec A + \tan A} \right)\left( {1 - ...

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

Explanation

Solution

In order to solve the question, we have to write all the terms such as tan⁡θ\tan \theta and sec⁡θ\sec \theta in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta .Use the algebraic identity (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} and trigonometric identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1, simplify the equation and get the answer.

Complete step-by-step answer:
Let consider T=(sec⁡A+tan⁡A)(1−sin⁡A)T = \left( {\sec A + \tan A} \right)\left( {1 - \sin A} \right)
Now we know that sec⁡A=1cos⁡A    &    tan⁡A=sin⁡Acos⁡A\sec A = \dfrac{1}{{\cos A}}\,\,\,\,\& \,\,\,\,\tan A = \dfrac{{\sin A}}{{\cos A}}
Using these conversions we get
T=(1cos⁡A+sin⁡Acos⁡A)(1−sin⁡A)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=(1+sin⁡A)(1−sin⁡A)cos⁡AT = \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 (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}in the numerator where a=1  ,  b=sin⁡Aa = 1\,\,,\,\,b = \sin A
T=1−sin⁡2Acos⁡AT = \dfrac{{1 - {{\sin }^2}A}}{{\cos A}}
Now we see 1  &  sin⁡2A1\,\,\& \,\,{\sin ^2}A in the numerator and we must check for the identity sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1. Rearranging this we get cos⁡2A=1−sin⁡2A{\cos ^2}A = 1 - {\sin ^2}A, using this in above equation we get
T=cos⁡2Acos⁡A=cos⁡AT = \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⁡θ\sin \theta and cos⁡θ\cos \theta at first.This question can also be solved by using the triangular trigonometric identities such as tan⁡θ=PB , sec⁡θ=HB , sin⁡θ=PH\tan \theta = \dfrac{P}{B}{\text{ , }}\sec \theta = \dfrac{H}{B}{\text{ , }}\sin\theta = \dfrac{P}{H} , where PP is the perpendicular, BB is the base and HH is the hypotenuse. Another property is also used P2+B2=H2{P^2} + {B^2} = {H^2} for this method.