MCQ
${\cot ^{ - 1}}\frac{3}{4} + {\sin ^{ - 1}}\frac{5}{{13}} = $
- ✓${\sin ^{ - 1}}\frac{{63}}{{65}}$
- B${\sin ^{ - 1}}\frac{{12}}{{13}}$
- C${\sin ^{ - 1}}\frac{{65}}{{68}}$
- D${\sin ^{ - 1}}\frac{5}{{12}}$
and $\sin \theta = \frac{1}{{\sqrt {1 + {{\cot }^2}\theta } }} = \frac{1}{{\sqrt {1 + (9/16)} }} = \frac{4}{5}$
Hence ${\cot ^{ - 1}}\frac{3}{4} + {\sin ^{ - 1}}\frac{5}{{13}} = {\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{5}{{13}}$
$ = {\sin ^{ - 1}}\left[ {\frac{4}{5}.\sqrt {1 - \frac{{25}}{{169}}} + \frac{5}{{13}}.\,\sqrt {1 - \frac{{16}}{{25}}} } \right]$
$ = {\sin ^{ - 1}}\left[ {\frac{4}{5}.\frac{{12}}{{13}} + \frac{5}{{13}}.\frac{3}{5}} \right]$
$ = {\sin ^{ - 1}}\left[ {\frac{{48 + 15}}{{65}}} \right] = {\sin ^{ - 1}}\frac{{63}}{{65}}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ The correct statement$(s)$ is(are)
$(A)$ $f^{\prime}(1) < 0$
$(B)$ $f(2) < 0$
$(C)$ $f^{\prime}(x) \neq 0$ for any $x \in(1,3)$
$(D)$ $f^{\prime}(x)=0$ for some $x \in(1,3)$
$2.$ If $\int_1^3 x^2 F^{\prime}(x) d x=-12$ and $\int_1^3 x^3 F^{\prime \prime}(x) d x=40$, then the correct expression$(s)$ is(are)
$(A)$ $9 f^{\prime}(3)+f^{\prime}(1)-32=0$
$(B)$ $\int_1^3 f(x) d x=12$
$(C)$ $9 f^{\prime}(3)-f^{\prime}(1)+32=0$
$(D)$ $\int_1^3 f(x) d x=-12$
Give the answer question $1$ and $2.$