Question

Find the solution for the expression \[\sec^2\left( {\tan^{ - 1}2} \right) + {\csc^2}\left( {\cot^{ - 1}3} \right)\]
A. 5
B. 13
C. 15
D. 6

Answer 1

Hint:Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Use the proper trigonometric formulae and identities including ratios to solve this equation.

Formulae used: \[{\tan ^{ - 1}}x = {{\mathop{\rm Sec}\nolimits} ^{ - 1}}\sqrt {1 + {x^2}} \]and \[\cot x =\csc{ - 1}[\sqrt {1 + {x^2}} ]\]

Complete Step by Step solution
Given: \[\sec^2\left[ {\tan^{ - 1}2} \right] + {\csc^2}\left[ {\cot^{ - 1}3} \right]\]
Taking the Squares common and using necessary trigonometric formulae we get,
\[\left [\sec\left( {\tan^{ - 1}2} \right) \right ]^{2} + \left [{\csc}\left( {\cot^{ - 1}3} \right) \right ]^{2}\]
Using the Trigonometric formulae \[{\tan ^{ - 1}}x = {\sec ^{ - 1}}\sqrt {1 + {x^2}} \]and \[\cot^{-1}x=\csc^{ - 1}[\sqrt {1 + {x^2}} ]\],
\[ = {\left[ {{\mathop{\rm Sec}\nolimits} [{{{\mathop{\rm Sec}\nolimits} }^{ - 1}}[\sqrt {1 + {2^2}} ]} \right]^2} + {\left[ {\csc[\cos e{c^{ - 1}}\sqrt {10} } \right]^2}\]
\[ = {\left[ {\sqrt 5 } \right]^2} + {\left[ {\sqrt {10} } \right]^2}\]
\[ = 5 + 10 = 15\]

Hence, the Solution is option C.

Additional information:
We can prove \[\tan^{-1}x=\sec^{-1}\sqrt{1+x^2}\] by using trigonometry identities.
Assume that, \[\tan^{-1}x = \theta\]
Taking tangent on both sides:
\[\Rightarrow \tan\left(\tan^{-1}x \right) = \tan\theta\]
\[\Rightarrow x = \tan\theta\]
We know that, \[ \sec^{2} \theta = 1+ \tan^{2} \theta\]
Substitute \[x = \tan\theta\] in the above equation:
\[ \sec^{2} \theta = 1+ x^{2}\]
Taking square root on both sides
\[ \sec \theta =\sqrt{ 1+ x^{2}}\]
\[\Rightarrow \theta =\sec^{-1}{\sqrt{ 1+ x^{2}}}\]
Since \[\tan^{-1}x = \theta\]
\[\Rightarrow \tan^{-1}x =\sec^{-1}{\sqrt{ 1+ x^{2}}}\]

Note: There are other ways to do this sum, where you can convert \[{\sec ^2}[{\tan ^{ - 1}}2]\]into \[1 + {\tan ^2}[{\tan ^{ - 1}}2]\]and \[\csc{^2}[{\cot ^{ - 1}}3]\] into \[1 + {\cot ^2}[{\cot ^{ - 1}}3]\]. Which I prefer to be an easier and simpler way to do. The student should be very well informed on trigonometric ratios and conversion formulae as well, both in inverse trigonometric functions and conversion formulae.