MCQ
${\sec ^2}({\tan ^{ - 1}}2) + {\rm{cose}}{{\rm{c}}^2}({\cot ^{ - 1}}3) = $
- A$5$
- B$13$
- ✓$15$
- D$6$
✓
Answer
Correct option: C.
$15$
c
(c) Let ${\tan ^{ - 1}}2 = \alpha \Rightarrow \tan \alpha = 2$
(c) Let ${\tan ^{ - 1}}2 = \alpha \Rightarrow \tan \alpha = 2$
and ${\cot ^{ - 1}}3 = \beta \Rightarrow \cot \beta = 3$
${\sec ^2}({\tan ^{ - 1}}2) + {\rm{cose}}{{\rm{c}}^{\rm{2}}}({\cot ^{ - 1}}3)$
$= {\sec ^2}\alpha + {\rm{cose}}{{\rm{c}}^{\rm{2}}}\alpha $
$=1 + {\tan ^2}\alpha + 1 + {\cot ^2}\alpha $
$= 2 + {(2)^2} + {(3)^2} = 15$.
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