যদি sinθ =4/5 এবং π/2 <  θ < π   হয়, তবে (tanθ+sec(-θ))/(cotθ+cosec(-θ) =কত?

Another Explanation (5):

প্রশ্নের সমাধান:

প্রদত্ত:

\(\sin \theta = \frac{4}{5}\) এবং \(\frac{\pi}{2} < \theta < \pi\)

ধাপ 1: \(\cos \theta\) নির্ণয়:

যেহেতু \(\sin^2 \theta + \cos^2 \theta = 1\), তাই:

\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \] এবং যেহেতু \(\theta\) দ্বিতীয় কোণের চতুর্দিকে (second quadrant), যেখানে \(\cos \theta < 0\), তাই: \[ \cos \theta = -\frac{3}{5} \]

ধাপ 2: \(\tan \theta\) নির্ণয়:

\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3} \]

ধাপ 3: \(\sec \theta\) নির্ণয়:

\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{3}{5}} = -\frac{5}{3} \]

ধাপ 4: \(\sec(-\theta)\):

\[ \sec(-\theta) = \sec \theta = -\frac{5}{3} \]

ধাপ 5: \(\cot \theta\):

\[ \cot \theta = \frac{1}{\tan \theta} = \frac{1}{-\frac{4}{3}} = -\frac{3}{4} \]

ধাপ 6: \(\csc(-\theta)\):

\[ \csc(-\theta) = -\csc \theta \] এবং, \[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{4}{5}} = \frac{5}{4} \] অতএব, \[ \csc(-\theta) = -\frac{5}{4} \]

ধাপ 7: মূল সমীকরণ:

\[ \frac{\tan \theta + \sec(-\theta)}{\cot \theta + \csc(-\theta)} = \frac{-\frac{4}{3} + (-\frac{5}{3})}{-\frac{3}{4} + (-\frac{5}{4})} \]

ধাপ 8: সাধারণ করা:

\[ = \frac{-\frac{4}{3} - \frac{5}{3}}{-\frac{3}{4} - \frac{5}{4}} = \frac{-\frac{9}{3}}{-\frac{8}{4}} = \frac{-3}{-2} = \frac{3}{2} \]

উত্তর:

\(\boxed{\frac{3}{2}}\)