
The value of $\left( \sec A+\tan A \right)\left( 1-\sin A \right)$ is equal to:
a). ${{\tan }^{2}}A$
b). ${{\sin }^{2}}A$
c). $\cos A$
d). $\sin A$
Answer
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Hint: Above question is a simple question of simplification of trigonometric function. So, to find the value of the above question we will just simplify $\left( \sec A+\tan A \right)\left( 1-\sin A \right)$ by writing $\sec A$ as $\dfrac{1}{\cos A}$ and $\tan A$ as $\dfrac{\sin A}{\cos A}$ and then solve them to get the required answer.
Complete step by step answer:
We can see that the above question is a simple question of simplification of trigonometric function. We know from the question that we have to find the value of $\left( \sec A+\tan A \right)\left( 1-\sin A \right)$. It means that we have to simplify $\left( \sec A+\tan A \right)\left( 1-\sin A \right)$ to get a simpler value as given in the option.
Since, we know that $\sec A=\dfrac{1}{\cos A}$ and $\tan A=\dfrac{\sin A}{\cos A}$ so, we will write them in place of $\sec A$ and $\tan A$ in $\left( \sec A+\tan A \right)\left( 1-\sin A \right)$ .
So, after replacing we will get:
$\left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)\left( 1-\sin A \right)$
Now, we will take $\cos A$ as LCM. Then, we can write it as:
$\left( \dfrac{1+\sin A}{\cos A} \right)\left( 1-\sin A \right)$
$=\dfrac{\left( 1+\sin A \right)\left( 1-\sin A \right)}{\cos A}$
$=\dfrac{\left( 1-{{\sin }^{2}}A \right)}{\cos A}$
Since, we know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ (identity of trigonometry )
So, we can say that ${{\cos }^{2}}A=1-{{\sin }^{2}}A$
Hence, $\dfrac{\left( 1-{{\sin }^{2}}A \right)}{\cos A}=\dfrac{{{\cos }^{2}}A}{\cos A}$
So, after cancelling cos A from numerator and denominator we will get:
$\dfrac{{{\cos }^{2}}A}{\cos A}=\cos A$
So, we will say that $\left( \sec A+\tan A \right)\left( 1-\sin A \right)=\cos A$.
So, the correct answer is “Option c”.
Note: Students are required to note that whenever we are asked to find the value of any trigonometric function then it means that we have to simplify that function as simply as possible if it is not a multiple-choice question. But if it is a multiple-choice question then we have to keep in mind the forms in which the options are given and then solve it so that we can find the value as per options given.
Complete step by step answer:
We can see that the above question is a simple question of simplification of trigonometric function. We know from the question that we have to find the value of $\left( \sec A+\tan A \right)\left( 1-\sin A \right)$. It means that we have to simplify $\left( \sec A+\tan A \right)\left( 1-\sin A \right)$ to get a simpler value as given in the option.
Since, we know that $\sec A=\dfrac{1}{\cos A}$ and $\tan A=\dfrac{\sin A}{\cos A}$ so, we will write them in place of $\sec A$ and $\tan A$ in $\left( \sec A+\tan A \right)\left( 1-\sin A \right)$ .
So, after replacing we will get:
$\left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)\left( 1-\sin A \right)$
Now, we will take $\cos A$ as LCM. Then, we can write it as:
$\left( \dfrac{1+\sin A}{\cos A} \right)\left( 1-\sin A \right)$
$=\dfrac{\left( 1+\sin A \right)\left( 1-\sin A \right)}{\cos A}$
$=\dfrac{\left( 1-{{\sin }^{2}}A \right)}{\cos A}$
Since, we know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ (identity of trigonometry )
So, we can say that ${{\cos }^{2}}A=1-{{\sin }^{2}}A$
Hence, $\dfrac{\left( 1-{{\sin }^{2}}A \right)}{\cos A}=\dfrac{{{\cos }^{2}}A}{\cos A}$
So, after cancelling cos A from numerator and denominator we will get:
$\dfrac{{{\cos }^{2}}A}{\cos A}=\cos A$
So, we will say that $\left( \sec A+\tan A \right)\left( 1-\sin A \right)=\cos A$.
So, the correct answer is “Option c”.
Note: Students are required to note that whenever we are asked to find the value of any trigonometric function then it means that we have to simplify that function as simply as possible if it is not a multiple-choice question. But if it is a multiple-choice question then we have to keep in mind the forms in which the options are given and then solve it so that we can find the value as per options given.
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