Simplify the following.

​$$\lparen\frac{tan\space40°}{cosec\space50°}\rparen^2+\lparen\frac{cot\space40°}{sec\space50°}\rparen^2$$​​

Correct option is D

Given:

Expression:

​$$\left(\frac{\tan 40^\circ}{\cosec 50^\circ}\right)^2 + \left(\frac{\cot 40^\circ}{\sec 50^\circ}\right)^2$$​​

Concept Used:
Use trigonometric identities and complementary angle relationships:

​$$\cosec(90^\circ - \theta) = \sec \theta$$​​

​$$\sec(90^\circ - \theta) = \cosec \theta$$​​

​$$\tan \theta = \frac{\sin \theta}{\cos \theta}, \cot \theta = \frac{\cos \theta}{\sin \theta}$$​​​

Solution:

$$\cosec 50^\circ = \frac{1}{\sin 50^\circ}, \sec 50^\circ = \frac{1}{\cos 50^\circ}$$​

​$$\left(\frac{\tan 40^\circ}{\cosec 50^\circ}\right)^2 = (\tan 40^\circ \cdot \sin 50^\circ)^2$$​​

​Now, $$\tan 40^\circ = \frac{\sin 40^\circ}{\cos 40^\circ}$$​, so:

​$$= \left(\frac{\sin 40^\circ}{\cos 40^\circ} \cdot \sin 50^\circ \right)^2$$​​

But $$\sin 50^\circ = \cos 40^\circ$$​, so:

​$$= \left(\frac{\sin 40^\circ}{\cos 40^\circ} \cdot \cos 40^\circ \right)^2 = (\sin 40^\circ)^2$$​​

Similarly,

​$$\left(\frac{\cot 40^\circ}{\sec 50^\circ}\right)^2 = \left(\cot 40^\circ \cdot \cos 50^\circ\right)^2$$​​

​$$\cot 40^\circ = \frac{\cos 40^\circ}{\sin 40^\circ}, \cos 50^\circ = \sin 40^\circ,$$​​

​$$= \left( \frac{\cos 40^\circ}{\sin 40^\circ} \cdot \sin 40^\circ \right)^2 = (\cos 40^\circ)^2$$​​

Now add both:

​$$(\sin 40^\circ)^2 + (\cos 40^\circ)^2 = 1$$​