Ex 14 pls how did they do this !

Q14. Find the value of sec²(tan^–13) + cosec² (cot^–12)

           We have to find sec² tan^–1 3 + cosec² (cot^–1 2)

          Now, sec tan^–1 3 = sec sec^–1 $\sqrt{1 +9}$ = sec sec^–1 $\sqrt{10} =\sqrt{10}$

           Also cosec (cot^–1 2) = cosec cosec^–1 $\sqrt{1 +2^2}$ = cosec cosec^–1 $\sqrt{5} =\sqrt{5}$

            Now, sec² tan^–1 3+ cosec²cot^–1 2 = ${\left(\sqrt{10}\right)}^2+{\left(\sqrt{5}\right)}^2 =10+5 =15$