Maths M.C.Q (1 Marks)

4. $\frac{7}{24}$

Solution:

 $\cot\Big[\cos^{-1}\Big(\frac{7}{25}\Big)\Big]$

$=\cot\Big(\cot^{-1}\frac{7}{24}\Big)$

$=\frac{7}{24}$

4. $ \frac {- \pi }{ 10 }$

Solution:

We have: $\sin ^{ -1 }{ \left( \cos { \frac { 53\pi }{ 5 } } \right) }=\sin ^{ -1 }{ \left( \cos { \frac { 50\pi+3\pi }{ 5 } } \right) }$

$ =\sin ^{ -1 }{ \left( \cos \left({ \frac { 50\pi}{5}+\frac{3\pi }{ 5 } }\right) \right) }$

$ =\sin ^{ -1 }{ \left( \cos \left({ 10\pi+\frac{3\pi }{ 5 } }\right) \right) }$

$ =\sin ^{ -1 }{ \left( \cos \left({ \frac{3\pi }{ 5 } }\right) \right) }, [\because \cos(2\text{n}\pi+\theta)=\cos\theta, n\in \text{Z}]$

$ =\sin ^{ -1 }{ \left( \sin \left({ \frac{\pi}{2}-\frac{3\pi }{ 5 } }\right) \right) },[∵\sin(2π​−θ)=\cosθ]$

$ =\sin ^{ -1 }{ \left( \sin \left(-\frac{\pi}{10 }\right) \right) } =−\frac{\pi}{10}$

Note $ \sin^{-1}(\sin \theta)=θ \text{ if} -\frac{\pi}{2}\le \theta\le \frac{\pi}{2}$