Inverse Trigonometric Functions Question 166
Question: $ {{\sec }^{2}}({{\tan }^{-1}}2)+cose{c^{2}}({{\cot }^{-1}}3)= $
[EAMCET 2001]
Options:
A) 5
B) 13
C) 15
D) 6
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ {{\tan }^{-1}}2=\alpha \Rightarrow \tan \alpha =2 $ and $ {{\cot }^{-1}}3=\beta \Rightarrow \cot \beta =3 $
$ {{\sec }^{2}}({{\tan }^{-1}}2)+cose{c^{2}}({{\cot }^{-1}}3) $ = $ {{\sec }^{2}}\alpha +cose{c^{2}}\alpha $ = $ 1+{{\tan }^{2}}\alpha +1+{{\cot }^{2}}\alpha $ = $ 2+{{(2)}^{2}}+{{(3)}^{2}}=15 $ .
BETA
