Question

Find the Solution of \[\left( {{\rm{sec }}2A + 1} \right){\sec ^2}A\].
A. \[\sec A\]
B. \[2\sec A\]
C. \[\sec 2A\]
D. \[2\sec 2A\]

Answer 1

Hint: Convert \[\sec 2A\] and \[{\sec ^2}A\] into their respective Cos forms and solve the equations. Use necessary exponential functions to further solve the equation to get the necessary answer

Formula Used: We use a trigonometric formula \[\cos 2x = 2{\cos ^2}x - 1\]

Complete step by step solution: Given expression is \[\left( {{\rm{sec }}2A + 1} \right){\sec ^2}A\]
Convert \[\sec 2A\] and \[{\sec ^2}A\] to \[\dfrac{1}{{\cos 2A}}\] and \[\dfrac{1}{{{{\cos }^2}A}}\]
 \[\left[ {\left( {\dfrac{1}{{\cos 2A}}} \right) + 1} \right]\left( {\dfrac{1}{{{{\cos }^2}A}}} \right)\]
 \[ = \left( {\dfrac{{1 + \cos 2A}}{{\cos 2A}}} \right)\left( {\dfrac{1}{{{{\cos }^2}A}}} \right)\]
Now, using formula \[\cos 2x = 2{\cos ^2}x - 1\],
Bringing \( - 1\) to RHS \[\cos 2x + 1 = 2{\cos ^2}x\]
\[ = \left( {\dfrac{{2{{\cos }^2}A}}{{\cos 2A}}} \right)\left( {\dfrac{1}{{{{\cos }^2}A}}} \right)\]
\[ = \dfrac{{2{{\cos }^2}A}}{{\cos 2A{{\cos }^2}A}}\]
As \[{\cos ^2}A\]cancels out in both Numerator and Denominator, we get,
\[ = \dfrac{2}{{\cos 2A}}\]
Then,
\[ = 2\left( {\dfrac{1}{{\cos 2A}}} \right)\]
Now converting \[\cos 2A\] into \[\sec 2A\]
\[ = 2\sec 2A\]

Option ‘D’ is correct

Additional Information: Trigonometry ratios deal with the ratios the length of sides of a right angle triangle. There are six ratios that are sin, cos, sec, cosec, tan, and cot. These trigonometry ratios are applicable for a right angle triangle. There are some identities in trigonometry. The identities are reciprocal trigonometric identities, Pythagorean trigonometric identities, ratio trigonometric identities, trigonometric identities of opposite angles, trigonometric identities of complementary angle, trigonometric identities of supplementary angle, sum and difference of angles trigonometric identities, double angle trigonometric identities, half angle trigonometric identities, product-sum trigonometric identities, trigonometric identities of product.



Note: Another way to solve this equation is instead of converting \[1 + \cos 2A\] into \[{\cos ^2}x\] , using the same formula we can convert \[{\cos ^2}x\] into \[\dfrac{{\cos 2A{\rm{ }} + 1}}{2}\] and then solve the equation to get the same answer.