
The value of $\left( {\sec 2A + 1} \right){\sec ^2}A$ is equal to:
A. $\sec A$
B. $2\sec A$
C. $\sec 2A$
D. $2\sec 2A$
Answer
495.9k+ views
Hint:The given question deals with finding the value of trigonometric expression doing basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as double angle formula for cosine ${\cos ^2}x - {\sin ^2}x = \cos 2x$. Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem. We must know the simplification rules to solve the problem with ease.
Complete step by step answer:
In the given problem, we have to find the value of $\left( {\sec 2A + 1} \right){\sec ^2}A$.
So, we have, $\left( {\sec 2A + 1} \right){\sec ^2}A$
Now, we know that secant function is the reciprocal of cosine function. So, we get,
$ \Rightarrow \left( {\dfrac{1}{{\cos 2A}} + 1} \right){\left( {\dfrac{1}{{\cos A}}} \right)^2}$
Using the trigonometric formula for double angle of cosine ${\cos ^2}x - {\sin ^2}x = \cos 2x$, we get,
$ \Rightarrow \left( {\dfrac{1}{{2{{\cos }^2}A - 1}} + 1} \right){\left( {\dfrac{1}{{\cos A}}} \right)^2}$
Taking LCM of the denominator, we get,
$ \Rightarrow \dfrac{{1 + 2{{\cos }^2}A - 1}}{{2{{\cos }^2}A - 1}} \times \dfrac{1}{{{{\cos }^2}A}}$
Adding up like terms, we get,
$ \Rightarrow \dfrac{{2{{\cos }^2}A}}{{2{{\cos }^2}A - 1}} \times \dfrac{1}{{{{\cos }^2}A}}$
Cancelling common terms in numerator and denominator, we get,
$ \Rightarrow \dfrac{2}{{2{{\cos }^2}A - 1}}$
Now, we know that $2{\cos ^2}x - 1 = \cos 2x$. Hence, we get,
$ \Rightarrow \dfrac{2}{{\cos 2A}}$
Secant and cosine functions are reciprocal of each other. So, we get,
$ \therefore 2\sec 2A$
So, we get the value of $\left( {\sec 2A + 1} \right){\sec ^2}A$ as $2\sec 2A$.
Hence, option D is the correct answer.
Note:We must have a strong grip over the concepts of trigonometry, related formulae and rules to ace these types of questions. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers but we have to mark the most appropriate option among the given choices. We must know the simplification rules to ease the calculations.
Complete step by step answer:
In the given problem, we have to find the value of $\left( {\sec 2A + 1} \right){\sec ^2}A$.
So, we have, $\left( {\sec 2A + 1} \right){\sec ^2}A$
Now, we know that secant function is the reciprocal of cosine function. So, we get,
$ \Rightarrow \left( {\dfrac{1}{{\cos 2A}} + 1} \right){\left( {\dfrac{1}{{\cos A}}} \right)^2}$
Using the trigonometric formula for double angle of cosine ${\cos ^2}x - {\sin ^2}x = \cos 2x$, we get,
$ \Rightarrow \left( {\dfrac{1}{{2{{\cos }^2}A - 1}} + 1} \right){\left( {\dfrac{1}{{\cos A}}} \right)^2}$
Taking LCM of the denominator, we get,
$ \Rightarrow \dfrac{{1 + 2{{\cos }^2}A - 1}}{{2{{\cos }^2}A - 1}} \times \dfrac{1}{{{{\cos }^2}A}}$
Adding up like terms, we get,
$ \Rightarrow \dfrac{{2{{\cos }^2}A}}{{2{{\cos }^2}A - 1}} \times \dfrac{1}{{{{\cos }^2}A}}$
Cancelling common terms in numerator and denominator, we get,
$ \Rightarrow \dfrac{2}{{2{{\cos }^2}A - 1}}$
Now, we know that $2{\cos ^2}x - 1 = \cos 2x$. Hence, we get,
$ \Rightarrow \dfrac{2}{{\cos 2A}}$
Secant and cosine functions are reciprocal of each other. So, we get,
$ \therefore 2\sec 2A$
So, we get the value of $\left( {\sec 2A + 1} \right){\sec ^2}A$ as $2\sec 2A$.
Hence, option D is the correct answer.
Note:We must have a strong grip over the concepts of trigonometry, related formulae and rules to ace these types of questions. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers but we have to mark the most appropriate option among the given choices. We must know the simplification rules to ease the calculations.
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