Question

cosec[2cot$^{-1}$(5) + cos$^{-1}$($\frac{4}{5}$)] is equal to :

• A. $\frac{56}{33}$
• B. $\frac{65}{33}$
• C. $\frac{65}{56}$
• D. $\frac{75}{56}$

Hint:

When dealing with inverse trigonometric functions, it's often easiest to convert them into standard trigonometric ratios by assigning them to an angle (e.g., $\alpha = \cot^{-1}(x)$) and then drawing a right-angled triangle to find the other ratios like sin($\alpha$) and cos($\alpha$).

Answer 1

The Correct Option is C

79. Solution: Let $\alpha=\cot^{-1}(5)$ and $\beta=\cos^{-1}\left(\frac{4}{5}\right)$. \[ \sin\alpha=\frac{1}{\sqrt{26}},\; \cos\alpha=\frac{5}{\sqrt{26}} \] \[ \sin\beta=\frac{3}{5},\; \cos\beta=\frac{4}{5} \] \[ \sin2\alpha=2\sin\alpha\cos\alpha=\frac{5}{13} \] \[ \cos2\alpha=\frac{12}{13} \] \[ \sin(2\alpha+\beta) =\frac{5}{13}\cdot\frac{4}{5} +\frac{12}{13}\cdot\frac{3}{5} =\frac{56}{65} \] \[ \csc(2\alpha+\beta)=\frac{65}{56} \] \[ \boxed{\frac{65}{56}} \]